Systems of Linear Equations and Word Problems

Note that we saw how to solve linear inequalities here in the Coordinate System and Graphing Lines section . Note also that we solve Algebra Word Problems without Systems here , and we solve systems using matrices in the Matrices and Solving Systems with Matrices  section here.

Introduction to Systems

“Systems of equations” just means that we are dealing with more than one equation and variable. So far, we’ve basically just played around with the equation for a line, which is $ y=mx+b$. Let’s say we have the following situation:

Now, you can always do “guess and check” to see what would work, but you might as well use algebra! It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.

The first trick in problems like this is to figure out what we want to know. This will help us decide what variables (unknowns) to use. What we want to know is how many pairs of jeans we want to buy (let’s say “$ j$”) and how many dresses we want to buy (let’s say “$ d$”). Always write down what your variables will be:

Let $ j=$ the number of jeans you will buy Let $ d=$ the number of dresses you’ll buy

Like we did before, let’s translate word-for-word from math to English:

Now we have the 2 equations as shown below. Notice that the $ j$ variable is just like the $ x$ variable and the $ d$ variable is just like the $ y$. It’s easier to put in $ j$   and $ d$ so we can remember what they stand for when we get the answers.

This is what we call a system, since we have to solve for more than one variable – we have to solve for 2 here. The cool thing is to solve for 2   variables , you typically need 2   equations , to solve for 3   variables , you need 3   equations , and so on. That’s easy to remember, right?

We need to get an answer that works in both equations ; this is what we’re doing when we’re solving; this is called solving simultaneous systems , or solving system simultaneously . There are several ways to solve systems; we’ll talk about graphing first.

Solving Systems by Graphing

Remember that when you graph a line, you see all the different coordinates (or $ x/y$ combinations) that make the equation work. In systems, you have to make both equations work, so the intersection of the two lines shows the point that fits both equations (assuming the lines do in fact intersect; we’ll talk about that later).  The points of intersections satisfy both equations simultaneously. 

Put these equations into the $ y=mx+b$ ($ d=mj+b$) format, by solving for the $ d$ (which is like the $ y$):

$ \displaystyle j+d=6;\text{ }\,\text{ }\text{solve for }d:\text{ }d=-j+6\text{ }$

$ \displaystyle 25j+50d=200;\text{ }\,\,\text{solve for }d:\text{ }d=\frac{{200-25j}}{{50}}=-\frac{1}{2}j+4$

Now graph both lines:

Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the  Exponents and Radicals in Algebra section. Also, t here are some examples of systems of inequality  here in the Coordinate System and Graphing Lines section .

Solving Systems with Substitution

Substitution is the favorite way to solve for many students! It involves exactly what it says: substituting one variable in another equation so that you only have one variable in that equation. Here is the same problem:

You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50 .  You really, really want to take home 6 items of clothing because you “need” that many new things. How many pairs of jeans and how many dresses you can buy so you use the whole $200 (tax not included)?

Below are our two equations, and let’s solve for “$ d$” in terms of “$ j$” in the first equation. Then, let’s substitute what we got for “$ d$” into the next equation. Even though it doesn’t matter which equation you start with, remember to always pick the “easiest” equation first (one that we can easily solve for a variable) to get a variable by itself.

We could buy 4 pairs of jeans and 2 dresses . Note that we could have also solved for “$ j$” first; it really doesn’t matter. You’ll want to pick the variable that’s most easily solved for. Let’s try another substitution problem that’s a little bit different:

Solving Systems with Linear Combination or Elimination

Probably the most useful way to solve systems is using linear combination, or linear elimination. The reason it’s most useful is that usually in real life we don’t have one variable in terms of another (in other words, a “$ y=$” situation).

The main purpose of the linear combination method is to add or subtract the equations so that one variable is eliminated. We can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below. We are using the Additive Property of Equality , Subtraction Property of Equality , Multiplicative Property of Equality , and/or Division Property of Equality that we saw here in the Types of Numbers and Algebraic Properties section :

If we have a set of 2 equations with 2 unknowns, for example, we can manipulate them by adding, multiplying or subtracting (we usually prefer adding) so that we get one equation with one variable. Let’s use our previous problem:

We could buy 4 pairs of jeans and 2 dresses .

Here’s another example:

Types of equations

In the example above, we found one unique solution to the set of equations. Sometimes, however, for a set of equations, there are no solutions (when lines are parallel) or an infinite number of solutions or infinitely many solutions (when the two lines are actually the same line, and one is just a “multiple” of the other).

When there is at least one solution , the equations are consistent equations , since they have a solution. When there is only one solution, the system is called independent , since they cross at only one point. When equations have infinite solutions, they are the same equation, are consistent , and are called dependent or coincident (think of one just sitting on top of the other).

When equations have no solutions , they are called inconsistent equations , since we can never get a solution . 

Here are graphs of inconsistent and dependent equations that were created on a graphing calculator:

Systems with Three Equations

Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns to solve for.

Let’s say at the same store, they also had pairs of shoes for $20 and we managed to get $60 more to spend! Now we have a new problem. To spend the even $260 , how many pairs of jeans, dresses, and pairs of shoes should we get if want, for example, exactly 10 total items (Remember that jeans cost $25 each and dresses cost $50 each).

Let’s let $ j=$ the number of pair of jeans, $ d=$ the number of dresses, and $ s=$ the number of pairs of shoes we should buy. So far, we’ll have the following equations:

$ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+\,20s=260\end{array}$

We’ll need another equation, since for three variables, we need three equations (otherwise, we theoretically might have infinite ways to solve the problem). In this type of problem, you would also need something like this: We want twice as many pairs of jeans as pairs of shoes . Now, since we have the same number of equations as variables , we can potentially get one solution for the system of equations. Here are the three equations:

We’ll learn later how to put these in our calculator to easily solve using matrices (see the  Matrices and Solving Systems with Matrices section). For now, we can use two equations at a time to eliminate a variable (using substitution and/or elimination), and keep doing this until we’ve solved for all variables. These can get really difficult to solve, but remember that in “real life”, there are computers to do all this work!

Remember again, that if we ever get to a point where we end up with something like this, it means there are an infinite number of solutions : $ 4=4$  (variables are gone and a number equals another number and they are the same). And if we up with something like this, it means there are no solutions : $ 5=2$ (variables are gone and two numbers are left and they don’t equal each other).

Let’s solve our system:      $ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+20s=260\\j=2s\end{array}$ :

We could buy 6 pairs of jeans, 1 dress, and 3 pairs of shoes .

Here’s one more example of a three-variable system of equations, where we’ll only use linear elimination:

$ \displaystyle \begin{align}5x-6y-\,7z\,&=\,7\\6x-4y+10z&=\,-34\\2x+4y-\,3z\,&=\,29\end{align}$

I know – this is really difficult stuff! But if you do it step-by-step and keep using the equations you need with the right variables, you can do it. Think of it like a puzzle – you may not know exactly where you’re going, but do what you can in baby steps, and you’ll get there (sort of like life sometimes, right?!). And we’ll learn much easier ways to do these types of problems.

Algebra Word Problems with Systems

Let’s do more word problems; you’ll notice that many of these are the same type that we did earlier in the Algebra Word Problems  section , but now we can use more than one variable. This will actually make the problems easier! Again, when doing these word problems:

  • If you’re wondering what the variables (or unknowns) should be when working on a word problem, look at what the problem is asking. These are usually (but not always) what your variables are!
  • If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!

Here are some problems:

Investment Word Problem

We also could have set up this problem with a table:

Mixture Word Problems

Here’s a mixture word problem . With mixture problems, remember if the problem calls for a pure solution or concentrate , use 100% (if the percentage is that solution) or 0% (if the percentage is another solution).

Let’s do the math (use substitution )!

$ \displaystyle \begin{array}{c}x\,\,+\,\,y=10\\.01x+.035y=10(.02)\end{array}$          $ \displaystyle \begin{array}{c}\,y=10-x\\.01x+.035(10-x)=.2\\.01x\,+\,.35\,\,-\,.035x=.2\\\,-.025x=-.15;\,\,\,\,\,x=6\\\,y=10-6=4\end{array}$

We would need 6 liters of the 1% milk, and 4 liters of the 3.5% milk.

Here’s another mixture problem:

$ \displaystyle \begin{array}{c}x+y=50\\8x+4y=50\left( {6.4} \right)\end{array}$                   $ \displaystyle \begin{array}{c}y=50-x\\8x+4\left( {50-x} \right)=320\\8x+200-4x=320\\4x=120\,;\,\,\,\,x=30\\y=50-30=20\\8x+4y=50(6.4)\end{array}$

We would need 30 pounds of the $8 coffee bean, and 20 pounds of the $4 coffee bean. See how similar this problem is to the one where we use percentages?

Distance Word Problem:

Here’s a distance word problem using systems ; distance problems have to do with an object’s speed, time, and distance. Note that, as well as the distance word problem here in the Algebra Word Problems section , there’s an example of a Parametric Distance Problem here in the Parametric Equations section .

Which Plumber Problem

Many word problems you’ll have to solve have to do with an initial charge or setup charge, and a charge or rate per time period. In these cases, the initial charge will be the $ \boldsymbol {y}$ -intercept , and the  rate will be the slope . Here is an example:

Geometry Word Problem:

Many times, we’ll have a geometry problem as an algebra word problem; these might involve perimeter, area, or sometimes angle measurements (so don’t forget these things!). Let’s do one involving angle measurements.

See – these are getting easier! Here’s one that’s a little tricky though:

Work Problem : 

Let’s do a “ work problem ” that is typically seen when studying Rational Equations (fraction with variables in them) and can be found here in the Rational Functions, E quations and Inequalities  section .

Note that there’s also a simpler version of this problem here in the Direct, Inverse, Joint and Combined Variation section .

Three Variable Word Problem:

Let’s do one more with three equations and three unknowns:

The “Candy” Problem

Sometimes we get lucky and can solve a system of equations where we have more unknowns (variables) then equations. (Actually, I think it’s not so much luck, but having good problem writers!) Here’s one like that:

There are more Systems Word Problems in the  Matrices and Solving Systems with Matrices section , Linear Programming section , and Right Triangle Trigonometry section .

Understand these problems, and practice, practice, practice!

For Practice : Use the Mathway  widget below to try a  Systems of Equations  problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve by Substitution or Solve by Addition/Elimination  to see the answer .

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps , or Click Here , you can register at Mathway for a free trial , and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Algebraic Functions, including Domain and Range   – you’re ready! 

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Solving Systems of Equations Real World Problems

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method . In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Now we are ready to apply these strategies to solve real world problems! Are you ready? First let's look at some guidelines for solving real world problems and then we'll look at a few examples.

Steps For Solving Real World Problems

  • Highlight the important information in the problem that will help write two equations.
  • Define your variables
  • Write two equations
  • Use one of the methods for solving systems of equations to solve.
  • Check your answers by substituting your ordered pair into the original equations.
  • Answer the questions in the real world problems. Always write your answer in complete sentences!

Ok... let's look at a few examples. Follow along with me. (Having a calculator will make it easier for you to follow along.)

Example 1: Systems Word Problems

You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs sold and the number of sodas sold. How many hot dogs were sold and how many sodas were sold?

1.  Let's start by identifying the important information:

  • hot dogs cost $1.50
  • Sodas cost $0.50
  • Made a total of $78.50
  • Sold 87 hot dogs and sodas combined

2.  Define your variables.

  • Ask yourself, "What am I trying to solve for? What don't I know?

In this problem, I don't know how many hot dogs or sodas were sold. So this is what each variable will stand for. (Usually the question at the end will give you this information).

Let x = the number of hot dogs sold

Let y = the number of sodas sold

3. Write two equations.

One equation will be related to the price and one equation will be related to the quantity (or number) of hot dogs and sodas sold.

1.50x + 0.50y = 78.50    (Equation related to cost)

 x + y = 87   (Equation related to the number sold)

4.  Solve! 

We can choose any method that we like to solve the system of equations. I am going to choose the substitution method since I can easily solve the 2nd equation for y.

Solving a systems using substitution

5. Think about what this solution means.

x is the number of hot dogs and x = 35. That means that 35 hot dogs were sold.

y is the number of sodas and y = 52. That means that 52 sodas were sold.

6.  Write your answer in a complete sentence.

35 hot dogs were sold and 52 sodas were sold.

7.  Check your work by substituting.

1.50x + 0.50y = 78.50

1.50(35) + 0.50(52) = 78.50

52.50 + 26 = 78.50

35 + 52 = 87

Since both equations check properly, we know that our answers are correct!

That wasn't too bad, was it? The hardest part is writing the equations. From there you already know the strategies for solving. Think carefully about what's happening in the problem when trying to write the two equations.

Example 2: Another Word Problem

You and a friend go to Tacos Galore for lunch. You order three soft tacos and three burritos and your total bill is $11.25. Your friend's bill is $10.00 for four soft tacos and two burritos. How much do soft tacos cost? How much do burritos cost?

  • 3 soft tacos + 3 burritos cost $11.25
  • 4 soft tacos + 2 burritos cost $10.00

In this problem, I don't know the price of the soft tacos or the price of the burritos.

Let x = the price of 1 soft taco

Let y = the price of 1 burrito

One equation will be related your lunch and one equation will be related to your friend's lunch.

3x + 3y = 11.25  (Equation representing your lunch)

4x + 2y = 10   (Equation representing your friend's lunch)

We can choose any method that we like to solve the system of equations. I am going to choose the combinations method.

Solving Systems Using Combinations

5. Think about what the solution means in context of the problem.

x = the price of 1 soft taco and x = 1.25.

That means that 1 soft tacos costs $1.25.

y = the price of 1 burrito and y = 2.5.

That means that 1 burrito costs $2.50.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

If you have difficulty with real world problems, you can find more examples and practice problems in the Algebra Class E-course.

Take a look at the questions that other students have submitted:

solving word problems with linear systems

Problem about the WNBA

Systems problem about ages

Problem about milk consumption in the U.S.

Vans and Buses? How many rode in each?

Telephone Plans problem

Systems problem about hats and scarves

Apples and guavas please!

How much did Alice spend on shoes?

All about stamps

Going to the movies

Small pitchers and large pitchers - how much will they hold?

Chickens and dogs in the farm yard

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Writing Systems of Linear Equations from Word Problems

As we learn about math, we often ask ourselves, "How is any of this knowledge useful in life? When will I ever need to use algebra?" Here's the thing: When you know how to "translate" word problems into algebraic equations, you'll immediately see how useful math really is. With a few simple steps, you can turn real questions about the world around you into equations, allowing you to calculate things that you never thought possible. It's one of the most interesting things about math -- so let's get started!

How to tell when a word problem can become a linear equation

First, we need to keep our eyes open for a number of clues. These clues tell us that we can turn our word problem into a linear equation:

  • There are different quantities of things, such as a specific number of people, objects, hours, and so on.
  • Each quantity has a clear value. Instead of saying, "There are a few boxes," the word problem needs to tell us how many boxes there are. Are there five? Six? Seven?
  • We need to know at least some of these values to build our linear equation. The unknowns can become variables, like "x" or "y."

How to turn word problems into linear equations

If we follow a few simple steps, we can turn certain word problems into linear equations:

  • Take a second to think about the "problem." What are we trying to find out? What is the "variable" in this word problem? Define all of the words carefully. If we can't define the words properly, our equation won't be accurate.
  • Turn the word problem into an equation. Plug in all of your known values and use letters like "x" and "y" to represent the unknown variables. Make sure you write out what each variable represents below the equation so you don't forget.
  • Solve the equation. Using our math skills, we can now solve the problem and find the values of our variables. We can use a wide range of strategies to solve the equation, including substitution, elimination, or graphing.

An example of a word problem translated into a linear equation

Now let's see how this all works with an example. Here's our word problem:

We decided to go to a music concert with all our friends, including 12 children and 3 adults. We paid for everyone's tickets for a total of $162. Another group of friends paid $122 for 8 children and 3 adults. How much does a child's ticket cost, and how much does an adult's ticket cost?

1. Understand the problem

We know two values: 12 children and 3 adults cost $162, while 8 children and 3 adults cost $122.

What we don't know is how much an adult's ticket costs, and how much a child's ticket costs.

Let's create variables for those unknowns:

x = the cost of one child's ticket

y = the cost of one adult's ticket

2. Translate the problem into an equation

We know that 12 children and 3 adults cost $162. Let's plug in our variables and create an equation based on this:

12x + 3y = 162

We can do the same for the other group of concert-goers:

8x + 3y = 122

3. Solve the equation

On one weekend they sold a total of 12 adult tickets and 3 child tickets for a total of 162 dollars, and the next weekend they sold 8 adult tickets and 3 child tickets for 122 dollars, find the price for a child's and adult's ticket.

When we have two very similar equations like this, we can simply subtract them from each other to get the values we need:

Now that we know the value of x, we can use it to find y.

12(10) + 3y = 162

120 + 3y = 162

Now we know that a child's ticket costs $10, while an adult's ticket costs $14.

We can now check our work by plugging our solutions back into our original equations:

12(10) + 3(14) = 162

8(10) + 3(14) = 122

You can use this strategy to solve all kinds of everyday math problems you encounter in life!

Topics related to the Writing Systems of Linear Equations from Word Problems

Consistent and Dependent Systems

Word Problems

Cramer's Rule

Flashcards covering the Writing Systems of Linear Equations from Word Problems

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Mathematics LibreTexts

3: Solving Linear Systems

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  • 3.1: Linear Systems with Two Variables and Their Solutions Real-world applications are often modeled using more than one variable and more than one equation. A system of equations consists of a set of two or more equations with the same variables. In this section, we will study linear systems consisting of two linear equations each with two variables.
  • 3.2: Solving Linear Systems with Two Variables In this section, we review a completely algebraic technique for solving systems, the substitution method11. The idea is to solve one equation for one of the variables and substitute the result into the other equation. After performing this substitution step, we are left with a single equation with one variable, which can be solved using algebra.
  • 3.3: Applications of Linear Systems with Two Variables If we translate an application to a mathematical setup using two variables, then we need to form a linear system with two equations. Setting up word problems with two variables often simplifies the entire process, particularly when the relationships between the variables are not so clear.
  • 3.4: Solving Linear Systems with Three Variables We can solve systems of three linear equations with three unknowns by elimination. If the process of solving a system leads to a false statement, then the system is inconsistent and has no solution. If the process of solving a system leads to a true statement, then the system is dependent and has infinitely many solutions.
  • 3.5: Matrices and Gaussian Elimination A linear system in upper triangular form can easily be solved using back substitution. The augmented coefficient matrix and Gaussian elimination can be used to streamline the process of solving linear systems.
  • 3.6: Determinants and Cramer’s Rule A square matrix is a matrix where the number of rows is the same as the number of columns. In this section we outline another method for solving linear systems using special properties of square matrices. Wen introduce the determinant  and show how  Cramer’s rule can be used to efficiently determine solutions to linear systems.
  • 3.7: Solving Systems of Inequalities with Two Variables A system of inequalities consists of a set of two or more inequalities with the same variables. The inequalities define the conditions that are to be considered simultaneously.
  • 3.E: Solving Linear Systems

RWM102: Algebra

solving word problems with linear systems

Solving Word Problems with Linear Systems

Read this article and watch the video. The article describes examples in which systems of equations can be used to solve real-world quantities. After you review, complete problems 1 to 4 and check your answers.

solving word problems with linear systems

Word Problems Linear Equations

Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician?

\(\textbf{1)}\) Joe and Steve are saving money. Joe starts with $105 and saves $5 per week. Steve starts with $5 and saves $15 per week. After how many weeks do they have the same amount of money? Show Equations \(y= 5x+105,\,\,\,y=15x+5\) Show Answer 10 weeks ($155)

\(\textbf{2)}\) mike and sarah collect rocks. together they collected 50 rocks. mike collected 10 more rocks than sarah. how many rocks did each of them collect show equations \(m+s=50,\,\,\,m=s+10\) show answer mike collected 30 rocks, sarah collected 20 rocks., \(\textbf{3)}\) in a classroom the ratio of boys to girls is 2:3. there are 25 students in the class. how many are girls show equations \(b+g=50,\,\,\,3b=2g\) show answer 15 girls (10 boys), \(\textbf{4)}\) kyle makes sandals at home. the sandal making tools cost $100 and he spends $10 on materials for each sandal. he sells each sandal for $30. how many sandals does he have to sell to break even show equations \(c=10x+100,\,\,\,r=30x\) show answer 5 sandals ($150), \(\textbf{5)}\) molly is throwing a beach party. she still needs to buy beach towels and beach balls. towels are $3 each and beachballs are $4 each. she bought 10 items in total and it cost $34. how many beach balls did she get show equations show answer 4 beachballs (6 towels), \(\textbf{6)}\) anna volunteers at a pet shelter. they have cats and dogs. there are 36 pets in total at the shelter, and the ratio of dogs to cats is 4:5. how many cats are at the shelter show equations \(c+d=40,\,\,\,5d=4c\) show answer 20 cats (16 dogs), \(\textbf{7)}\) a store sells oranges and apples. oranges cost $1.00 each and apples cost $2.00 each. in the first sale of the day, 15 fruits were sold in total, and the price was $25. how many of each type of frust was sold show equations \(o+a=15,\,\,\,1o+2a=25\) show answer 10 apples and 5 oranges, \(\textbf{8)}\) the ratio of red marbles to green marbles is 2:7. there are 36 marbles in total. how many are red show equations \(r+g=36,\,\,\,7r=2g\) show answer 8 red marbles (28 green marbles), \(\textbf{9)}\) a tennis club charges $100 to join the club and $10 for every hour using the courts. write an equation to express the cost \(c\) in terms of \(h\) hours playing tennis. show equation the equation is \(c=10h+100\), \(\textbf{10)}\) emma and liam are saving money. emma starts with $80 and saves $10 per week. liam starts with $120 and saves $6 per week. after how many weeks will they have the same amount of money show equations \(e = 10x + 80,\,\,\,l = 6x + 120\) show answer 10 weeks ($180 each), \(\textbf{11)}\) mark and lisa collect stamps. together they collected 200 stamps. mark collected 40 more stamps than lisa. how many stamps did each of them collect show equations \(m + l = 200,\,\,\,m = l + 40\) show answer mark collected 120 stamps, lisa collected 80 stamps., \(\textbf{12)}\) in a classroom, the ratio of boys to girls is 3:5. there are 40 students in the class. how many are boys show equations \(b + g = 40,\,\,\,5b = 3g\) show answer 15 boys (25 girls), \(\textbf{13)}\) lisa is selling handmade jewelry. the materials cost $60, and she sells each piece for $20. how many pieces does she have to sell to break even show equations \(c=60,\,\,\,r=20x\) show answer 3 pieces, \(\textbf{14)}\) tom is buying books and notebooks for school. books cost $15 each, and notebooks cost $3 each. he bought 12 items in total, and it cost $120. how many notebooks did he buy show equations \(b + n = 12,\,\,\,15b+3n=120\) show answer 5 notebooks (7 books), \(\textbf{15)}\) emily volunteers at an animal shelter. they have rabbits and guinea pigs. there are 36 animals in total at the shelter, and the ratio of guinea pigs to rabbits is 4:5. how many guinea pigs are at the shelter show equations \(r + g = 36,\,\,\,5g=4r\) show answer 16 guinea pigs (20 rabbits), \(\textbf{16)}\) mike and sarah are going to a theme park. mike’s ticket costs $40, and sarah’s ticket costs $30. they also bought $20 worth of food. how much did they spend in total show equations \(m + s + f = t,\,\,\,m=40,\,\,\,s=30,\,\,\,f=20\) show answer they spent $90 in total., \(\textbf{17)}\) the ratio of red marbles to blue marbles is 2:3. there are 50 marbles in total. how many are blue show equations \(r + b = 50,\,\,\,3r=2b\) show answer 30 blue marbles (20 red marbles), \(\textbf{18)}\) a pizza restaurant charges $12 for a large pizza and $8 for a small pizza. if a customer buys 5 pizzas in total, and it costs $52, how many large pizzas did they buy show equations \(l + s = 5,\,\,\,12l+8s=52\) show answer they bought 3 large pizzas (2 small pizzas)., \(\textbf{19)}\) the area of a rectangle is 48 square meters. if the length is 8 meters, what is the width of the rectangle show equations \(a=l\times w,\,\,\,l=8,\,\,\,a=48\) show answer the width is 6 meters., \(\textbf{20)}\) two numbers have a sum of 50. one number is 10 more than the other. what are the two numbers show equations \(x+y=50,\,\,\,x=y+10\) show answer the numbers are 30 and 20., \(\textbf{21)}\) a store sells jeans for $40 each and t-shirts for $20 each. in the first sale of the day, they sold 8 items in total, and the price was $260. how many of each type of item was sold show equations \(j+t=8,\,\,\,40j+20t=260\) show answer 5 jeans and 3 t-shirts were sold., \(\textbf{22)}\) the ratio of apples to carrots is 3:4. there are 28 fruits in total. how many are apples show equations \(\)a+c=28,\,\,\,4a=3c show answer there are 12 apples and 16 carrots., \(\textbf{23)}\) a phone plan costs $30 per month, and there is an additional charge of $0.10 per minute for calls. write an equation to express the cost \(c\) in terms of \(m\) minutes. show equation the equation is \(\)c=30+0.10m, \(\textbf{24)}\) a triangle has a base of 8 inches and a height of 6 inches. calculate its area. show equations \(a=0.5\times b\times h,\,\,\,b=8,\,\,\,h=6\) show answer the area is 24 square inches., \(\textbf{25)}\) a store sells shirts for $25 each and pants for $45 each. in the first sale of the day, 4 items were sold, and the price was $180. how many of each type of item was sold show equations \(t+p=4,\,\,\,25t+45p=180\) show answer 0 shirts and 4 pants were sold., \(\textbf{26)}\) a garden has a length of 12 feet and a width of 10 feet. calculate its area. show equations \(a=l\times w,\,\,\,l=12,\,\,\,w=10\) show answer the area is 120 square feet., \(\textbf{27)}\) the sum of two consecutive odd numbers is 56. what are the two numbers show equations \(x+y=56,\,\,\,x=y+2\) show answer the numbers are 27 and 29., \(\textbf{28)}\) a toy store sells action figures for $15 each and toy cars for $5 each. in the first sale of the day, 10 items were sold, and the price was $110. how many of each type of item was sold show equations \(a+c=10,\,\,\,15a+5c=110\) show answer 6 action figures and 4 toy cars were sold., \(\textbf{29)}\) a bakery sells pie for $2 each and cookies for $1 each. in the first sale of the day, 14 items were sold, and the price was $25. how many of each type of item was sold show equations \(p+c=14,\,\,\,2p+c=25\) show answer 11 pies and 3 cookies were sold., \(\textbf{for 30-33}\) two car rental companies charge the following values for x miles. car rental a: \(y=3x+150 \,\,\) car rental b: \(y=4x+100\), \(\textbf{30)}\) which rental company has a higher initial fee show answer company a has a higher initial fee, \(\textbf{31)}\) which rental company has a higher mileage fee show answer company b has a higher mileage fee, \(\textbf{32)}\) for how many driven miles is the cost of the two companies the same show answer the companies cost the same if you drive 50 miles., \(\textbf{33)}\) what does the \(3\) mean in the equation for company a show answer for company a, the cost increases by $3 per mile driven., \(\textbf{34)}\) what does the \(100\) mean in the equation for company b show answer for company b, the initial cost (0 miles driven) is $100., \(\textbf{for 35-39}\) andy is going to go for a drive. the formula below tells how many gallons of gas he has in his car after m miles. \(g=12-\frac{m}{18}\), \(\textbf{35)}\) what does the \(12\) in the equation represent show answer andy has \(12\) gallons in his car when he starts his drive., \(\textbf{36)}\) what does the \(18\) in the equation represent show answer it takes \(18\) miles to use up \(1\) gallon of gas., \(\textbf{37)}\) how many miles until he runs out of gas show answer the answer is \(216\) miles, \(\textbf{38)}\) how many gallons of gas does he have after 90 miles show answer the answer is \(7\) gallons, \(\textbf{39)}\) when he has \(3\) gallons remaining, how far has he driven show answer the answer is \(162\) miles, \(\textbf{for 40-42}\) joe sells paintings. each month he makes no commission on the first $5,000 he sells but then makes a 10% commission on the rest., \(\textbf{40)}\) find the equation of how much money x joe needs to sell to earn y dollars per month. show answer the answer is \(y=.1(x-5,000)\), \(\textbf{41)}\) how much does joe need to sell to earn $10,000 in a month. show answer the answer is \($105,000\), \(\textbf{42)}\) how much does joe earn if he sells $45,000 in a month show answer the answer is \($4,000\), see related pages\(\), \(\bullet\text{ word problems- linear equations}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- averages}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- consecutive integers}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- distance, rate and time}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- break even}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- ratios}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- age}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- mixtures and concentration}\) \(\,\,\,\,\,\,\,\,\), linear equations are a type of equation that has a linear relationship between two variables, and they can often be used to solve word problems. in order to solve a word problem involving a linear equation, you will need to identify the variables in the problem and determine the relationship between them. this usually involves setting up an equation (or equations) using the given information and then solving for the unknown variables . linear equations are commonly used in real-life situations to model and analyze relationships between different quantities. for example, you might use a linear equation to model the relationship between the cost of a product and the number of units sold, or the relationship between the distance traveled and the time it takes to travel that distance. linear equations are typically covered in a high school algebra class. these types of problems can be challenging for students who are new to algebra, but they are an important foundation for more advanced math concepts. one common mistake that students make when solving word problems involving linear equations is failing to set up the problem correctly. it’s important to carefully read the problem and identify all of the relevant information, as well as any given equations or formulas that you might need to use. other related topics involving linear equations include graphing and solving systems. understanding linear equations is also useful for applications in fields such as economics, engineering, and physics., about andymath.com, andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. if you have any requests for additional content, please contact andy at [email protected] . he will promptly add the content. topics cover elementary math , middle school , algebra , geometry , algebra 2/pre-calculus/trig , calculus and probability/statistics . in the future, i hope to add physics and linear algebra content. visit me on youtube , tiktok , instagram and facebook . andymath content has a unique approach to presenting mathematics. the clear explanations, strong visuals mixed with dry humor regularly get millions of views. we are open to collaborations of all types, please contact andy at [email protected] for all enquiries. to offer financial support, visit my patreon page. let’s help students understand the math way of thinking thank you for visiting. how exciting.

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System-of-Equations Word Problems

Exercises More Exercises

Many problems lend themselves to being solved with systems of linear equations. In "real life", these problems can be incredibly complex. This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics.

In your studies, however, you will generally be faced with much simpler problems. What follows are some typical examples.

The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?

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System of Equations Word Problems

In the past , I would have set this up by picking a variable for one of the groups (say, " c " for "children") and then use "(total) less (what I've already accounted for)" (in this case, " 2200 –  c ") for the other group. Using a system of equations, however, allows me to use two different variables for the two different unknowns.

number of adults: a

number of children: c

With these variables, I can create equations for the totals they've given me:

total number: a + c = 2200

total income: 4 a + 1.5 c = 5050

Now I can solve the system for the number of adults and the number of children. I will solve the first equation for one of the variables, and then substitute the result into the other equation:

a = 2200 – c

4(2200 – c ) + 1.5 c = 5050

8800 – 4 c + 1.5 c = 5050

8800 – 2.5 c = 5050

–2.5 c = –3750

Now I can back-solve for the value of the other variable:

a = 2200 – (1500) = 700

I have values for my two variables. I can look back at my definitions for the variables to interpret these values. To answer the original question, there were:

1500 children and 700 adults.

You will probably start out with problems which, like the one above, seem very familiar. But you will then move on to more complicated problems.

The sum of the digits of a two-digit number is 7 . When the digits are reversed, the number is increased by 27 . Find the number.

The trick here is to work with the digits explicitly. I'll use " t " for the "tens" digit of the original number and " u " for the "units" (or "ones") digit. I then have:

The ten's digit stands for "ten times of this digit's value". Just as "26" is "10 times 2, plus 6 times 1", so also the two-digit number they've given me will be ten times the "tens" digit, plus one times the "units" digit. In other words:

original number: 10 t + 1 u

The new number has the values of the digits (represented by the variables) in reverse order. This gives me:

new number: 10 u + 1 t

And this new number is twenty-seven more than the original number. The keyword "is" means "equals", so I get:

(new number) is (old number) increased by (twenty-seven)

10 u + 1 t = (10 t + 1 u ) + 27

Now I have a system of equations that I can solve:

10 u + t = 10 t + u + 27

First I'll simplify the second equation:

9 u – 9 t = 27

u – t = 3

After reordering the variables in the first equation, I now have:

Adding down , I get:

Then t = 2 . Back-solving, this means that the original number was 25 and the new number (gotten by switching the digits) is 52 . Since 52 – 25 = 27 , this solution checks out.

The number is 25 .

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Find the equation of the parabola that passes through the points (–1, 9) , (1, 5) , and (2, 12) .

Recalling that a parabola has a quadratic as its equation, I know that I am looking for an equation of the form ax 2  +  bx  +  c = y . Also, I know that points are of the form ( x ,  y ) . Practically speaking, this mean that, in each of these points, they have given me values for x and y that make the quadratic equation true. Plugging the three points in the general equation for a quadratic, I get a system of three equations, where the variables stand for the unknown coefficients of that quadratic:

a (–1) 2 + b (–1) + c = 9

a (1) 2 + b (1) + c = 5

a (2) 2 + b (2) + c = 12

Simplifying the three equations, I get:

1 a – b + c = 9

1 a + b + c = 5

4 a + 2 b + c = 12

I won't display the solving of this problem, but the result is that a  = 3, b  = –2, and c  = 4 , so the equation they're wanting is:

y = 3 x 2 – 2 x + 4

You may also see similar exercises referring to circles, using:

x 2 + y 2 + bx + cy + d = 0

...or other conics, though parabolas are the most common. Keep in mind that projectile problems (like shooting an arrow up in the air or dropping a penny from the roof of a tall building) are also parabola problems, using:

–( 1 / 2 ) gt 2 + v 0 t + h 0 = s

...where h 0 is the original height, v 0 is the initial velocity, s is the height at time t , usually measured in seconds, and g refers to gravity, being 9.8 if you're working in meters and 32 if you're working in feet).

All of these different permutations of the above example work the same way: Take the general equation for the curve, plug in the given points, and solve the resulting system of equations for the values of the coefficients. Warning: If you see an exercise of this sort in the homework, be advised that you may be expected to know the forms of the general equations (such as " ax 2 + bx + c = y " for parabolas) on the next text.

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solving word problems with linear systems

Mathx

  • Linear systems – word problems

When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. Once you do that, these linear systems are solvable just like other linear systems . The same rules apply. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you accustomed to finding elements of linear systems inside of word problems. Example 1 : A farmhouse shelters 16 animals. Some of them are chickens and the others are cows. Altogether these animals have 60 legs. How many chickens and how many cows are in the farmhouse?

First, to make the calculations clearer, we will choose symbols to represent the number of cows and the number of chickens. Let us say that the chickens will be represented with x and the cows with y. Now, this task gave us enough information to make two equations. The first one is that the sum of the number of chickens (x) and the number of cows (y) is 16, since there are only 16 animals in the farmhouse. That equation should look like this: x + y = 16

The second piece of information we have is that the total number of legs in the farmhouse is 60. Since we know that cows have four legs each and chickens have two legs each, we have enough information to make another equation. This one will look like this: 2*x + 4*y = 60

Now we have a system of linear equations with two equations and two variables. The only thing left to do now is to solve the system. We will solve it here for you, but if you need to remind yourself how to do that step by step, read the article called Systems of linear equations . x = 16 – y 32 – 2y + 4y = 60 2y = 28 y = 14 x = 2

We can now see that there are two chickens and 14 cows in the farmhouse. The next example will be a bit harder,

Example 2: Rodney’s Kitchen Supplies makes and sells spoons and forks. It costs the store $2 to buy the supplies needed to make a fork, and $1 for the supplies needed to make a spoon. The store sells the forks for $4 and the spoons for 5$. Last month Rodney’s Kitchen Supplies spent $39 on supplies and sold the all of the forks and spoons that were made last month using those supplies for $93. How many forks and spoons did they make?

As we did in the first example, we will first designate symbols to available variables. So, the number of forks made will be represented with x and the number of spoons with y. Again, we have enough information to make two equations. The total cost of making a particular number of forks (x), which cost $2 to make each, and a particular number of spoons (y), which cost $1 to make each, is $39. So that will be our first equation and it will look like this: 2*x + y = 39 The other piece of information tells us that if we sell that number of forks (x) for $4 each and that number of spoons (y) for $5 each, we will make $93. And that will be our second equation: 4*x + 5*y = 93 This was the hard part. Now all we have to do is to solve this linear system to find how many spoons and how many forks did we make last month. y = 39 – 2x 4x + 5*(39 – 2x) = 93 4x + 195 – 10x = 93 -6x = 93 – 195 -6x = -102 |: (-6) x = 17 y = 5

We can see that last month the store made and sold 17 forks and five spoons.

linear systems problems

Although they can seem complicated, mastery and understanding of linear systems and associated word problems will come with a bit of practice. With experience you will be able to recognize their elements and solve even complicated systems with ease. Feel free to use the math worksheets below to practice solving this type of linear systems.

Linear systems – word problems exams for teachers

Linear systems – word problems worksheets for studets.

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SAT Math Word Problems

Most students tremble with fear when they think about preparing for the Math portion of the SAT .  After all, there are 20+ topics tested, and the test requires mastery of topics ranging from arithmetic to trigonometry. Some topics cover basic math questions from algebra class, such as solving linear equations or recognizing graphs of parabolas. Solving questions such as these is tough enough, but the SAT adds a layer of difficulty by presenting word problems from a wide range of disciplines that require you to use your algebra skills to solve them.

To tackle these word problems, you must first understand the context of the word problem and then create the equations that you’ll need to solve, based on the information provided in the question.

In this article, I’ll first cover two critical algebra skills you need to master. Then I’ll provide a number of strategies for efficiently solving hard SAT word problems, as well as examples, so we can see these strategies in action and avoid common mistakes.

Here are the topics we’ll cover:

Example 1: linear equation with one variable, sales tax: example 1, substitution method practice 1, substitution method practice 2, when dealing with word problems, translation is key, balancing equations: example 1, balancing equations: example 2, balancing equations: example 3, “number of items”: example 1, “number of items”: example 2, age problem: example 1, money problem: example 1, in conclusion: sat math word problems, what’s next.

We’ll start with solving basic linear equations, then we’ll review the substitution method of solving SAT algebra problems. Finally, we’ll use these two techniques to solve math word problem examples.

Basic Algebra – Solving Single-Variable Linear Equations 

The simplest algebra questions on the SAT are single-variable linear equations. Your goal is to solve for the value of the variable:

  • 2v + 5 = 3v – 7

Note that solving single-variable equations generally requires that we combine all terms containing the variable on one side of the equation and all constant terms on the other side.

Let’s practice simplifying and solving an equation with one variable.

If 8q – 2 = 10q + 14, then what is the value of q?

First, we combine our constant terms by adding 2 to both sides:

8q – 2 + 2 = 10q + 14 + 2

8q = 10q + 16

Next, we combine the variable terms by subtracting 10q from both sides of the equation:

8q – 10q = 10q + 16 – 10q

Finally, we divide both sides of the equation by -2 to isolate the variable:

-2q / -2 = 16 / -2

When solving for a single variable, we combine like terms and isolate the variable we are solving for.

Now, let’s practice with an example.

If -2x – 11 = 5x + 3, then x equals which of the following?

-2x – 11 = 5x + 3

-2x = 5x + 14

Now let’s turn our attention to applying algebra to solve a straightforward word problem involving sales tax.

Sales Tax Questions

A common single-variable word problem that you might be asked on the SAT involves the calculation of sales tax. When you buy an item, you not only pay for the item itself but you also generally pay a surcharge called sales tax, typically stated as a percentage of the cost of the item. We add the sales tax to the cost of the item to determine the actual amount we will pay at the register.

The equation for purchasing an item in an area where sales tax is charged is:

Cost of Item + Sales Tax = Total

Let’s look at a basic example of the calculation of sales tax.

Laura buys a blender for $40.00 in a town where the sales tax rate is 4.5%. How much will she have to pay, including sales tax?

We see that the cost of the item is $40.00, and the sales tax rate is 4.5%. Thus, the sales tax amount is 4.5% of $40.00, or 40 x 0.045. Let’s put these into the equation:

40 + (40)(0.045) = Total

40 + 1.8 = Total

41.8 = Total

Thus, Laura will pay a total of $41.80: the item cost of $40.00 and the sales tax of $1.80.

Know the sales tax formula: Cost of Item + Sales Tax = Total

Let’s look at a sample SAT question about sales tax.

Alex buys a pair of shoes. The sales tax rate in his town is 6%. The register total for his purchase, including sales tax, is $115.54. How much sales tax did Alex pay?

We set up the linear equation by noting that the total purchase price is equal to the cost of the item plus the sales tax:

Total = Cost of Item + Sales Tax

We are given the total and the sales tax rate. However, we don’t know the actual cost of the item, so we will let x = the cost of the item:

115.54 = x + (0.06)(x)

We combine like terms on the right side of the equation and solve for x:

115.54 = 1.06x

115.54 / 1.06 = x

We see that the cost of the item is $109.00. Now we calculate the sales tax:

109 x 0.06 = 6.54

The sales tax is $6.54.

Next, let’s discuss linear equations with two variables.

Linear Equations with Two Variables/Systems of Linear Equations

Two-variable equations have not just one but two different variables:

  • v – u = 12
  • 2x + y = 10
  • 5z – 3y = 42

These are called two-variable equations because they contain two different variables . If we have a second equation that contains one or both of the variables, the two equations are called a system of linear equations. One common way to solve for the values of the variables is by using the substitution method . Let’s discuss the substitution method now.

Using the Substitution Method to Solve a System of Linear Equations

When we’re working with SAT systems of equations word problems, the essence of the substitution method is that we first isolate one variable in one of the equations. Then we substitute whatever that variable is equal to into the other equation. Let’s practice with an example.

In the system of linear equations below, determine the value of a.

b = 3a (equation 1)

a + b = 12 (equation 2)

Looking at our two equations, we see that b is already isolated in equation 1. So, we can substitute 3a (from equation one) for b (in equation two). This gives us:

a + 3a = 12

For the system of linear equations below, what is the value of x?

3x + y = 11 (equation 1)

2x + 5y = 3 (equation 2)

First, let’s isolate y in equation 1 by subtracting 3x from both sides of the equation:

y = 11 – 3x

Now, we can substitute 11 – 3x for y in equation 2. This gives us:

2x + 5(11 – 3x) = 3

2x + 55 – 15x = 3

We can use the substitution method to solve a system of linear equations.

Next, let’s discuss the many topics on the SAT in which we use algebra to determine the solution.

Here are a couple of sample SAT math word problems that you might encounter:

Luca buys apples, which cost $2 each, and bananas, which cost $3 each, at the market. If he buys twice as many bananas as apples and spends $48 at the market, how many bananas does he buy?

Tyrone and Greg have a total of 40 marbles. If Tyrone has 4 times the number of marbles that Greg has, how many marbles does Greg have?

The bottom line is that a word problem presents a scenario requiring us to translate the given information into an algebraic equation that we then solve.

Word problems are not just about solving equations; they are also about translating words into equations! Let’s look at some common translations:

  • “Is” translates to equals ( = )

Kendra is the same age as Carla

Kendra’s age = Carla’s age

  • “More” translates to addition ( + )

Arita has 6 more marbles than Pablo

Arita = Pablo + 6

  • “Less/fewer” translates to subtraction (-)

Sammy has 3 fewer coins than Rati

Sammy = Rati – 3

  • “Times as many” translates to multiplication (✖)

Harold has 5 times as many newspapers as Phoebe

Harold = Phoebe ✖ 5

Know the common translations of words to algebraic equations.

Before jumping into word problem practice questions, let’s discuss one point of confusion students have when translating words into equations: properly balancing the equations.

You Must Make Sure Your Equations are Balanced

A basic principle for translating words into equations is that the equations must be balanced correctly. If they are not, you might obtain reversed values for your variables’ values, or you might get a negative answer. Let’s look at a few correct and incorrect ways to balance equations, and we’ll explain why the equations are balanced correctly.

Sherry has 30 more dollars than Melanie.

When we are setting up this equation, we must understand that Sherry has more money than Melanie. For example, if Melanie has 20 dollars, then Sherry has 50 dollars.

So, to properly balance the equation, we must add 30 dollars to Melanie’s amount and set it equal to Sherry’s amount.

If we let S = Sherry’s money and M = Melanie’s money, we have:

Nala has 4 fewer toys than Frank.

When considering this equation, we must understand that Nala has fewer toys than Frank.

In other words, Frank has more toys than Nala.

So, to set up a balanced equation, we must subtract 4 from Frank’s amount and set that equal to Nala’s.

If we let N = the number of Nala’s toys and F = the number of Frank’s toys, we have:

N – 4 = F

There are 5 times as many baseballs as tennis balls.

When considering this equation, we must understand that there are more baseballs than tennis balls.

So, to set up a correct equation, we multiply the number of tennis balls by 5 and set that amount equal to the number of baseballs.

If we let B = the number of baseballs and T = the number of tennis balls, we have:

Now that we are familiar with how to translate, balance, and solve equations, let’s jump into the various types of general word problems you will encounter on the SAT, and how to solve them.

“Number of Items” Questions

We have already encountered some simple examples of “number of items” questions in this article. Now, let’s attack a full-fledged problem like those you might encounter on test day.

Julia and Tory have a total of 30 candy bars. If Julia has 5 times as many candy bars as Tory, how many candy bars does Tory have?

First, we define our variables.

Let’s let J = the number of candy bars Julia has and T = the number of candy bars Tory has.

Next, let’s create equations based on the information given in the question stem.

“Julia and Tory have a total of 30 candy bars” is translated as:

J + T = 30 (equation 1)

“Julia has 5 times as many candy bars as Tory” is translated as:

J = 5T (equation 2)

Next, we use the substitution method by substituting 5T for J in equation 1:

5T + T = 30

Let’s practice one more. This time, we will add a twist.

Leti has 4 times as many cookies as Henrik. If Leti gives Henrik 7 cookies, Henrik will have 4 fewer cookies than Leti. How many cookies did Leti originally have?

We can let L = the number of cookies Leti has and H = the number of cookies Henrik has. Next, we use the information from the question stem to create equations.

Because Leti has 4 times as many cookies as Henrik, we have:

L = 4H (equation 1)

After Leti gives Henrik 7 cookies, Leti will have L – 7 cookies. After Henrik gets the 7 cookies from Leti, he will have H + 7 cookies. He will still have 4 fewer cookies than Leti.

Thus, our second equation is:

L – 7 = (H + 7) +4

L = H + 18 (equation 2)

Next, we can use the substitution method, substituting 4H for L in equation 2:

4H = H + 18

Henrik originally had 6 cookies. Thus, Leti originally had 4 x 6 = 24 cookies.

Next, let’s discuss another common type of SAT word problem: age problems.

Age Problems

Age problems are a common type of SAT problem with a unique spin. Instead of having a straightforward translation, as we practiced above, an age problem will usually make us translate words into equations based on an age in the past, an age in the future, or even both.

For example, we might need to consider Ann’s age 10 years from now. To do this, we let Ann’s age today = A, so her age in 10 years will be A + 10.

Similarly, Ann’s age 12 years ago would be expressed as A – 12.

Age problems generally require us to compare ages in the past or future.

Chantal is 6 years younger than Yolanda. If, in 5 years, Yolanda will be twice as old as Chantal, how old is Chantal today?

Let’s let C = Chantal’s age today and Y = Yolanda’s age today. Next, let’s create some equations.

Because Chantal is 6 years younger than Yolanda, we have:

C = Y – 6 (equation 1)

Now we express what each age will be in 5 years:Chantal will be C + 5 and Yolanda will be Y + 6. The relationship between their ages in 5 years is that Yolanda will be twice Chantal’s age:

Y + 5 = 2(C + 5)

Y + 5 = 2C + 10 (equation 2)

Next, we use the substitution method and substitute Y – 6 for C in equation 2, and we have:

Y + 5 = 2(Y – 6) + 10

Y + 5 = 2Y – 12 + 10

Y + 5 = 2Y – 2

Yolanda is 7 years old today. Thus, Chantal is 7 – 6 = 1 year old today.

Next, let’s discuss money problems.

Money Problems

In general, money problems on the SAT involve two different commodities, such as adult tickets and child tickets, and the revenue earned from selling them. We can create two equations, one relating the number of items and the second relating the revenue earned from the sale of the items.

We will see that the substitution method is extremely useful for solving the two equations that are created from the given information.

Be prepared to use the substitution method when dealing with money problems.

At an ice cream stand, customers can buy either a small cone for $3.00 or a large cone for $4.50. Yesterday, the owner reported he had sold 720 cones, and his revenue for the day was $2,340. How many small cones did he sell?

First, let’s define our variables. We can let the number of small cones sold = x and the number of large cones sold = y.

Next, let’s create our equations. Because the owner sold 720 ice cream cones, we have:

x + y = 720 (equation 1)

His total revenue was $2,340 from selling x small cones for $3.00 each and y large cones for $4.50 each:

3x + 4.5y = 2,340 (equation 2)

Let’s isolate x in equation 1:

x = 720 – y

Now, let’s substitute 720 – y for x into equation 2 and solve:

3(720 – y) + 4.5y = 2,340

2,160 – 3y + 4.5y = 2,340

Since the number of large cones sold is 120, the number of small cones sold is 720 – 120 = 600.

Of the 20+ major math topics on the SAT, one of the most challenging is word problems. In order to be successful with this type of problem, you need to be skilled both at translating words into equations and at solving algebra equations. The ability to solve SAT word problems is a must for scoring well on the exam.

The two most common skills for solving math word problems are knowing how to solve a linear equation and using the substitution method for solving systems of linear equations. You must translate the question into an equation or equations first, then use the appropriate algebraic method to arrive at the answer.

Some of the common types of word problems that we have covered here are sales tax questions, “number of items” questions, age questions, and money questions.

In this article, we have covered how to solve word problems on the SAT. You have a good start in doing well on this particular topic. But don’t lose your perspective: word problems are only one of many topics you must master in order to get a great SAT score.  Read our article on how to improve your SAT score for more strategies, tips, and tricks.

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Jeffrey Miller is the head SAT instructor for Target Test Prep. Jeff has more than 14 years of experience in the business of helping students with low SAT scores hurdle the seemingly impossible and achieve the scores they need to get into the top 20 colleges in the world. Jeff has cultivated many successful college graduates through his SAT instruction, and will be a pivotal resource for many more to follow.

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Students’ Polya problem solving skills on system of linear equations with two variables bases on mathematical ability

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Pathuddin , Dewi Hamidah , Silvana Panggalo , Zun Azizul Hakim , Muhammad Suwardin , Bakri , Evie Awuy; Students’ Polya problem solving skills on system of linear equations with two variables bases on mathematical ability. AIP Conf. Proc. 16 February 2024; 3046 (1): 020018. https://doi.org/10.1063/5.0194577

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This study seeks students’ abilities to solve Polya problems involving systems of linear equations with two variables based on their mathematical prowess. Three junior high school students with high, average, and low math aptitudes were chosen as participants. Research data were collected using a written test on the system of linear equations with two variables problem in the form of word questions and interviews. The time triangulation technique is used to establish the credibility of the research data. The findings demonstrated that students with strong mathematical aptitudes could comprehend problems, plan solutions, and carry out problem-solving strategies. Still, they lacked the knowledge necessary to confirm the accuracy of the outcomes, leading students to doubt the proper solution. Students with moderate math abilities can only understand problems, plan solutions and carry out problem-solving methods. Students won’t believe the correct answer since they don’t know how to double-check the results. Meanwhile, students with low math ability can only understand the problem and fail to make problem-solving plans. Students with weak math skills make problem-solving mistakes due to this incompetence. The research’s findings can be used to create learning models that will motivate students to develop their ability to solve mathematical problems, particularly those involving the system of linear equations with two variables material, by emphasizing the accuracy of information conversion into variables, understanding of the fundamental ideas underlying the use of procedures, and the significance of double-checking the outcomes of system of linear equations with two variables problem-solving.

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