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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
Algebra 1 Flashcards
College Algebra Flashcards
Practice tests covering the Writing Systems of Linear Equations from Word Problems
Algebra 1 Diagnostic Tests
College Algebra Diagnostic Tests
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To solve a system of linear equations word problem: Select variables to represent the unknown quantities. Using the given information, write a system of two linear equations relating the two variables. Solve the system of linear equations using either substitution or elimination. Let's look at an example!
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Expert Answer. Let the three numbers be x,y and z x+y+z = 119----1 z = 3 …. = Solving a word problem using a 3x3 system of linear equations.... The sum of three numbers is 119. The third number is 3 times the second. The second number is 6 more than the first.
Sal solves a word problem about the angles of a given triangle by modeling the given information as a system of three equations and variables. Created by Sal Khan and Monterey Institute for Technology and Education. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted cezagara 8 years ago
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?
Solution: The number is denoted by x .8 less than x mean, that we need to subtract 8 from it. We get: x − 8 For example, 8 less than 10 is 10 − 8 = 2. d) Subtract 5 p 2 − 7 p + 2 from 3 p 2 + 4 p and simplify. Solution: We need to calculate 3 p 2 + 4 p minus 5 p 2 − 7 p + 2: ( 3 p 2 + 4 p) − ( 5 p 2 − 7 p + 2) Simplifying this expression gives:
Solving algebraic word problems requires us to combine our ability to create equations and solve them. To solve an algebraic word problem: Define a variable. Write an equation using the variable. Solve the equation. If the variable is not the answer to the word problem, use the variable to calculate the answer.
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This video walks through an example of solving a word problem:(i) defining relevant variables,(ii) writing a system of linear equations with three variables,...
David Severin. All Sal is doing in this video is finding the key features of a linear equation given in slope intercept (almost) form y=mx+b. Given Q=15+0.4t, we can switch to slope intercept form by moving right side to Q=0.4t+15 noting the y is represented by Q and x is represented by t. The slope is 0.4 and the y intercept is 15.
How to solve a word problem with a 3 by 3 system of linear equations
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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.
Solving a word problem using a 3x3 system of linear equations: Problem type 1
example 1: Solve by using Gaussian elimination: example 2: Solve by using Cramer's rule About Cramer's rule This calculator uses Cramer's rule to solve systems of three equations with three unknowns. The Cramer's rule can be stated as follows: Given the system: with then the solution of this system is:
Solving a word problem using a 3x3 system of linear equations: Problem type 1
Systems of equations; Tips for entering queries. Enter your queries using plain English. To avoid ambiguous queries, make sure to use parentheses where necessary. Here are some examples illustrating how to ask about solving systems of equations. solve y = 2x, y = x + 10; solve system of equations {y = 2x, y = x + 10, 2x = 5y} y = x^2 - 2, y = 2 ...
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