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solving linear equations sample problems

solving linear equations sample problems

1.7 Solving Linear Equations

Learning objectives.

  • Use the properties of equality to solve basic linear equations.
  • Identify and solve conditional linear equations, identities, and contradictions.
  • Clear fractions from equations.
  • Set up and solve linear applications.

Solving Basic Linear Equations

An equation Statement indicating that two algebraic expressions are equal. is a statement indicating that two algebraic expressions are equal. A linear equation with one variable An equation that can be written in the standard form a x + b = 0 , where a and b are real numbers and a ≠ 0 . , x , is an equation that can be written in the standard form a x + b = 0 where a and b are real numbers and a ≠ 0 . For example,

3 x − 12 = 0

A solution Any value that can replace the variable in an equation to produce a true statement. to a linear equation is any value that can replace the variable to produce a true statement. The variable in the linear equation 3 x − 12 = 0 is x and the solution is x = 4 . To verify this, substitute the value 4 in for x and check that you obtain a true statement.

3 x − 12 = 0 3 ( 4 ) − 12 = 0 12 − 12 = 0 0 = 0         ✓

Alternatively, when an equation is equal to a constant, we may verify a solution by substituting the value in for the variable and showing that the result is equal to that constant. In this sense, we say that solutions “satisfy the equation.”

Is a = − 1 2 a solution to − 10 a + 5 = 25 ?

Recall that when evaluating expressions, it is a good practice to first replace all variables with parentheses, and then substitute the appropriate values. By making use of parentheses, we avoid some common errors when working the order of operations.

− 10 a + 5 = − 10 ( − 1 2 ) + 5 = 5 + 5 = 10 ≠ 25         ✗

Answer: No, a = − 1 2 does not satisfy the equation.

Developing techniques for solving various algebraic equations is one of our main goals in algebra. This section reviews the basic techniques used for solving linear equations with one variable. We begin by defining equivalent equations Equations with the same solution set. as equations with the same solution set.

3 x − 5 = 16                 3 x = 21                 x = 7 }               E q u i v a l e n t   e q u a t i o n s

Here we can see that the three linear equations are equivalent because they share the same solution set, namely, {7}. To obtain equivalent equations, use the following properties of equality Properties that allow us to obtain equivalent equations by adding, subtracting, multiplying, and dividing both sides of an equation by nonzero real numbers. . Given algebraic expressions A and B , where c is a nonzero number:

Note: Multiplying or dividing both sides of an equation by 0 is carefully avoided. Dividing by 0 is undefined and multiplying both sides by 0 results in the equation 0 = 0.

We solve algebraic equations by isolating the variable with a coefficient of 1. If given a linear equation of the form a x + b = c , then we can solve it in two steps. First, use the appropriate equality property of addition or subtraction to isolate the variable term. Next, isolate the variable using the equality property of multiplication or division. Checking the solution in the following examples is left to the reader.

Solve: 7 x − 2 = 19 .

7 x − 2 = 19 7 x − 2   +   2 = 19   +   2                   A d d   2   t o   b o t h   s i d e s .   7 x = 21 7 x 7 = 21 7                           D i v i d e   b o t h   s i d e s   b y   7 . x = 3

Answer: The solution is 3.

Solve: 56 = 8 + 12 y .

When no sign precedes the term, it is understood to be positive. In other words, think of this as 56 = + 8 + 12 y . Therefore, we begin by subtracting 8 on both sides of the equal sign.

56 − 8 = 8 + 12 y − 8 48 = 12 y 48 12 = 12 y 12 4 = y

It does not matter on which side we choose to isolate the variable because the symmetric property Allows you to solve for the variable on either side of the equal sign, because x = 5 is equivalent to 5 = x . states that 4 = y is equivalent to y = 4 .

Answer: The solution is 4.

Solve: 5 3 x + 2 = − 8 .

Isolate the variable term using the addition property of equality, and then multiply both sides of the equation by the reciprocal of the coefficient 5 3 .

5 3 x + 2 = − 8 5 3 x + 2 − 2 = − 8 − 2                                     S u b t r a c t   2   o n   b o t h   s i d e s . 5 3 x = − 10 3 5 ⋅ 5 3 x = 3 5 ⋅ ( − 10 ) − 2                     M u l t i p l y   b o t h   s i d e s   b y   3 5 . 1 x = 3 ⋅ ( − 2 ) x = − 6

Answer: The solution is −6.

In summary, to retain equivalent equations, we must perform the same operation on both sides of the equation.

Try this! Solve: 2 3 x + 1 2 = − 5 6 .

Answer: x = − 2

General Guidelines for Solving Linear Equations

Typically linear equations are not given in standard form, and so solving them requires additional steps. When solving linear equations, the goal is to determine what value, if any, will produce a true statement when substituted in the original equation. Do this by isolating the variable using the following steps:

  • Step 1: Simplify both sides of the equation using the order of operations and combine all like terms on the same side of the equal sign.
  • Step 2: Use the appropriate properties of equality to combine like terms on opposite sides of the equal sign. The goal is to obtain the variable term on one side of the equation and the constant term on the other.
  • Step 3: Divide or multiply as needed to isolate the variable.
  • Step 4: Check to see if the answer solves the original equation.

We will often encounter linear equations where the expressions on each side of the equal sign can be simplified. If this is the case, then it is best to simplify each side first before solving. Normally this involves combining same-side like terms.

Note: At this point in our study of algebra the use of the properties of equality should seem routine. Therefore, displaying these steps in this text, usually in blue, becomes optional.

Solve: − 4 a + 2 − a = 1 .

First combine the like terms on the left side of the equal sign.

− 4 a + 2 − a = 1                             C o m b i n e   s a m e - s i d e   l i k e   t e r m s . − 5 a + 2 = 1                                   S u b t r a c t   2   o n   b o t h   s i d e s . − 5 a = − 1                             D i v i d e   b o t h   s i d e s   b y   − 5. a = − 1 − 5 = 1 5

Always use the original equation to check to see if the solution is correct.

− 4 a + 2 − a = − 4 ( 1 5 ) + 2 − 1 5 = − 4 5 + 2 1 ⋅ 5 5 − 1 5 = − 4 + 10 + 1 5 = 5 5 = 1             ✓

Answer: The solution is 1 5 .

Given a linear equation in the form a x + b = c x + d , we begin the solving process by combining like terms on opposite sides of the equal sign. To do this, use the addition or subtraction property of equality to place like terms on the same side so that they can be combined. In the examples that remain, the check is left to the reader.

Solve: − 2 y − 3 = 5 y + 11 .

Subtract 5 y on both sides so that we can combine the terms involving y on the left side.

− 2 y − 3 − 5 y = 5 y + 11 − 5 y − 7 y − 3 = 11

From here, solve using the techniques developed previously.

− 7 y − 3 = 11                           A d d   3   t o   b o t h   s i d e s . − 7 y   =   14 y = 14 − 7                       D i v i d e   b o t h   s i d e s   b y   − 7. y   =   − 2

Answer: The solution is −2.

Solving will often require the application of the distributive property.

Solve: − 1 2 ( 10 x − 2 ) + 3 = 7 ( 1 − 2 x ) .

Simplify the linear expressions on either side of the equal sign first.

− 1 2 ( 10 x − 2 ) + 3 = 7 ( 1 − 2 x )            D i s t r i b u t e . − 5 x + 1 + 3 = 7 − 14 x                   C o m b i n e   s a m e - s i d e   l i k e   t e r m s . − 5 x + 4 = 7 − 14 x                   C o m b i n e   o p p o s i t e - s i d e   l i k e   t e r m s . 9 x = 3                                           S o l v e . x = 3 9 = 1 3

Answer: The solution is 1 3 .

Solve: 5 ( 3 − a ) − 2 ( 5 − 2 a ) = 3 .

Begin by applying the distributive property.

5 ( 3 − a ) − 2 ( 5 − 2 a ) = 3 15 − 5 a − 10 + 4 a = 3 5 − a = 3 − a = − 2

Here we point out that − a is equivalent to − 1 a ; therefore, we choose to divide both sides of the equation by −1.

− a = − 2 − 1 a − 1 = − 2 − 1 a = 2

Alternatively, we can multiply both sides of − a = − 2 by negative one and achieve the same result.

− a = − 2 ( − 1 ) ( − a ) = ( − 1 ) ( − 2 ) a = 2

Answer: The solution is 2.

Try this! Solve: 6 − 3 ( 4 x − 1 ) = 4 x − 7 .

Answer: x = 1

There are three different types of equations. Up to this point, we have been solving conditional equations Equations that are true for particular values. . These are equations that are true for particular values. An identity An equation that is true for all possible values. is an equation that is true for all possible values of the variable. For example, x   =   x             I d e n t i t y has a solution set consisting of all real numbers, ℝ . A contradiction An equation that is never true and has no solution. is an equation that is never true and thus has no solutions. For example, x   +   1   =   x                   C o n t r a d i c t i o n has no solution. We use the empty set, Ø , to indicate that there are no solutions.

If the end result of solving an equation is a true statement, like 0 = 0, then the equation is an identity and any real number is a solution. If solving results in a false statement, like 0 = 1, then the equation is a contradiction and there is no solution.

Solve: 4 ( x + 5 ) + 6 = 2 ( 2 x + 3 ) .

4 ( x + 5 ) + 6 = 2 ( 2 x + 3 ) 4 x + 20 + 6 = 4 x + 6 4 x + 26 = 4 x + 6 26 = 6       ✗

Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.

Solve: 3 ( 3 y + 5 ) + 5 = 10 ( y + 2 ) − y .

3 ( 3 y + 5 ) + 5 = 10 ( y + 2 ) − y 9 y + 15 + 5 = 10 y + 20 − y 9 y + 20 = 9 y + 20 9 y = 9 y 0 = 0             ✓

Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.

The coefficients of linear equations may be any real number, even decimals and fractions. When this is the case it is possible to use the multiplication property of equality to clear the fractional coefficients and obtain integer coefficients in a single step. If given fractional coefficients, then multiply both sides of the equation by the least common multiple of the denominators (LCD).

Solve: 1 3 x + 1 5 = 1 5 x − 1 .

Clear the fractions by multiplying both sides by the least common multiple of the given denominators. In this case, it is the L C D ( 3 , 5 ) = 15 .

15 ⋅ ( 1 3 x + 1 5 ) = 15 ⋅ ( 1 5 x − 1 )                             M u l t i p l y   b o t h   s i d e s   b y   1 5 . 15 ⋅ 1 3 x + 15 ⋅ 1 5 = 15 ⋅ 1 5 x − 15 ⋅ 1               S i m p l i f y . 5 x + 3 = 3 x − 15                                                 S o l v e . 2 x = − 18 x = − 18 2 = − 9

Answer: The solution is −9.

It is important to know that this technique only works for equations. Do not try to clear fractions when simplifying expressions. As a reminder:

We simplify expressions and solve equations. If you multiply an expression by 6, you will change the problem. However, if you multiply both sides of an equation by 6, you obtain an equivalent equation.

Applications Involving Linear Equations

Algebra simplifies the process of solving real-world problems. This is done by using letters to represent unknowns, restating problems in the form of equations, and by offering systematic techniques for solving those equations. To solve problems using algebra, first translate the wording of the problem into mathematical statements that describe the relationships between the given information and the unknowns. Usually, this translation to mathematical statements is the difficult step in the process. The key to the translation is to carefully read the problem and identify certain key words and phrases.

When translating sentences into mathematical statements, be sure to read the sentence several times and parse out the key words and phrases. It is important to first identify the variable, “ let x represent… ” and state in words what the unknown quantity is. This step not only makes our work more readable, but also forces us to think about what we are looking for.

When 6 is subtracted from twice the sum of a number and 8 the result is 5. Find the number.

Let n represent the unknown number.

solving linear equations sample problems

To understand why we included the parentheses in the set up, you must study the structure of the following two sentences and their translations:

The key was to focus on the phrase “ twice the sum ,” this prompted us to group the sum within parentheses and then multiply by 2. After translating the sentence into a mathematical statement we then solve.

2 ( n + 8 ) − 6 = 5 2 n + 16 − 6 = 5 2 n + 10 = 5 2 n = − 5 n = − 5 2

2 ( n + 8 ) − 6 = 2 ( − 5 2 + 8 ) − 6 = 2 ( 11 2 ) − 6 = 11 − 6 = 5                           ✓

Answer: The number is − 5 2 .

General guidelines for setting up and solving word problems follow.

  • Step 1: Read the problem several times, identify the key words and phrases, and organize the given information.
  • Step 2: Identify the variables by assigning a letter or expression to the unknown quantities.
  • Step 3: Translate and set up an algebraic equation that models the problem.
  • Step 4: Solve the resulting algebraic equation.
  • Step 5: Finally, answer the question in sentence form and make sure it makes sense (check it).

For now, set up all of your equations using only one variable. Avoid two variables by looking for a relationship between the unknowns.

A rectangle has a perimeter measuring 92 meters. The length is 2 meters less than 3 times the width. Find the dimensions of the rectangle.

The sentence “ The length is 2 meters less than 3 times the width ,” gives us the relationship between the two variables.

Let w represent the width of the rectangle.

Let 3 w − 2 represent the length.

solving linear equations sample problems

The sentence “ A rectangle has a perimeter measuring 92 meters ” suggests an algebraic set up. Substitute 92 for the perimeter and the expression 3 w − 2 for the length into the appropriate formula as follows:

P = 2 l + 2 w ↓   ↓ 92 = 2 ( 3 w − 2 ) + 2 w

Once you have set up an algebraic equation with one variable, solve for the width, w .

92 = 2 ( 3 w − 2 ) + 2 w D i s t r i b u t e . 92 = 6 w − 4 + 2 w C o m b i n e   l i k e   t e r m s . 92 = 8 w − 4 S o l v e   f o r   w . 96 = 8 w 12 = w

Use 3 w − 2 to find the length.

l = 3 w − 2 = 3 ( 12 ) − 2 = 36 − 2 = 34

To check, make sure the perimeter is 92 meters.

P = 2 l         +       2 w = 2 ( 34 ) + 2 ( 12 ) = 68 + 24 = 92

Answer: The rectangle measures 12 meters by 34 meters.

Given a 4 3 8 % annual interest rate, how long will it take $2,500 to yield $437.50 in simple interest?

Let t represent the time needed to earn $437.50 at 4 3 8 % . Organize the information needed to use the formula for simple interest, I = p r t .

Next, substitute all of the known quantities into the formula and then solve for the only unknown, t .

I = p r t 437.50 = 2500 ( 0.04375 ) t 437.50 = 109.375 t 437.50 109.375 = 109.375 t 109.375 4 = t

Answer: It takes 4 years for $2,500 invested at 4 3 8 % to earn $437.50 in simple interest.

Susan invested her total savings of $12,500 in two accounts earning simple interest. Her mutual fund account earned 7% last year and her CD earned 4.5%. If her total interest for the year was $670, how much was in each account?

The relationship between the two unknowns is that they total $12,500. When a total is involved, a common technique used to avoid two variables is to represent the second unknown as the difference of the total and the first unknown.

     Let x represent the amount invested in the mutual fund.

     Let 12,500 − x represent the remaining amount invested in the CD.

     Organize the data.

The total interest is the sum of the interest earned from each account.

m u t u a l   f u n d   i n t e r e s t                 +                         C D     i n t e r e s t                     =           t o t a l   i n t e r e s t           0.07 x                   +                 0.045 ( 12,500 − x )               =                       670

This equation models the problem with one variable. Solve for x .

0.07 x + 0.045 ( 12,500 − x )     = 670 0.07 x + 562.5 − 0.045 x = 670 0.025 x + 562.5 = 670 0.025 x = 107.5 x = 107.5 0.025 x = 4,300

Use 12 , 500 − x to find the amount in the CD.

12,500 − x = 12,500 − 4,300 = 8,200

Answer: Susan invested $4,300 at 7% in a mutual fund and $8,200 at 4.5% in a CD.

Key Takeaways

  • Solving general linear equations involves isolating the variable, with coefficient 1, on one side of the equal sign. To do this, first use the appropriate equality property of addition or subtraction to isolate the variable term on one side of the equal sign. Next, isolate the variable using the equality property of multiplication or division. Finally, check to verify that your solution solves the original equation.
  • If solving a linear equation leads to a true statement like 0 = 0, then the equation is an identity and the solution set consists of all real numbers, ℝ .
  • If solving a linear equation leads to a false statement like 0 = 5, then the equation is a contradiction and there is no solution, Ø .
  • Clear fractions by multiplying both sides of an equation by the least common multiple of all the denominators. Distribute and multiply all terms by the LCD to obtain an equivalent equation with integer coefficients.
  • Simplify the process of solving real-world problems by creating mathematical models that describe the relationship between unknowns. Use algebra to solve the resulting equations.

Topic Exercises

Part a: solving basic linear equations.

Determine whether or not the given value is a solution.

− 5 x + 4 = − 1 ;    x = − 1

4 x − 3 = − 7 ;    x = − 1

3 y − 4 = 5 ;    y = 9 3

− 2 y + 7 = 12 ;    y = − 5 2

3 a − 6 = 18 − a ;    a = − 3

5 ( 2 t − 1 ) = 2 − t ;    t = 2

a x − b = 0 ;    x = b a

a x + b = 2 b ;    x = b a

5 x − 3 = 27

6 x − 7 = 47

4 x + 13 = 35

6 x − 9 = 18

9 a + 10 = 10

5 − 3 a = 5

− 8 t + 5 = 15

− 9 t + 12 = 33

  • 2 3 x + 1 2 = 1
  • 3 8 x + 5 4 = 3 2
  • 1 − 3 y 5 = 2
  • 2 − 5 y 6 = − 8

Solve for x : a x − b = c

Solve for x : a x + b = 0

Part B: Solving Linear Equations

6 x − 5 + 2 x = 19

7 − 2 x + 9 = 24

12 x − 2 − 9 x = 5 x + 8

16 − 3 x − 22 = 8 − 4 x

5 y − 6 − 9 y = 3 − 2 y + 8

7 − 9 y + 12 = 3 y + 11 − 11 y

3 + 3 a − 11 = 5 a − 8 − 2 a

2 − 3 a = 5 a + 7 − 8 a

1 3 x − 3 2 + 5 2 x = 5 6 x + 1 4

5 8 + 1 5 x − 3 4 = 3 10 x − 1 4

1.2 x − 0.5 − 2.6 x = 2 − 2.4 x

1.59 − 3.87 x = 3.48 − 4.1 x − 0.51

5 − 10 x = 2 x + 8 − 12 x

8 x − 3 − 3 x = 5 x − 3

5 ( y + 2 ) = 3 ( 2 y − 1 ) + 10

7 ( y − 3 ) = 4 ( 2 y + 1 ) − 21

7 − 5 ( 3 t − 9 ) = 22

10 − 5 ( 3 t + 7 ) = 20

5 − 2 x = 4 − 2 ( x − 4 )

2 ( 4 x − 5 ) + 7 x = 5 ( 3 x − 2 )

4 ( 4 a − 1 ) = 5 ( a − 3 ) + 2 ( a − 2 )

6 ( 2 b − 1 ) + 24 b = 8 ( 3 b − 1 )

2 3 ( x + 18 ) + 2 = 1 3 x − 13

2 5 x − 1 2 ( 6 x − 3 ) = 4 3

1.2 ( 2 x + 1 ) + 0.6 x = 4 x

6 + 0.5 ( 7 x − 5 ) = 2.5 x + 0.3

5 ( y + 3 ) = 15 ( y + 1 ) − 10 y

3 ( 4 − y ) − 2 ( y + 7 ) = − 5 y

1 5 ( 2 a + 3 ) − 1 2 = 1 3 a + 1 10

3 2 a = 3 4 ( 1 + 2 a ) − 1 5 ( a + 5 )

6 − 3 ( 7 x + 1 ) = 7 ( 4 − 3 x )

6 ( x − 6 ) − 3 ( 2 x − 9 ) = − 9

3 4 ( y − 2 ) + 2 3 ( 2 y + 3 ) = 3

5 4 − 1 2 ( 4 y − 3 ) = 2 5 ( y − 1 )

− 2 ( 3 x + 1 ) − ( x − 3 ) = − 7 x + 1

6 ( 2 x + 1 ) − ( 10 x + 9 ) = 0

Solve for w : P = 2 l + 2 w

Solve for a : P = a + b + c

Solve for t : D = r t

Solve for w : V = l w h

Solve for b : A = 1 2 b h

Solve for a : s = 1 2 a t 2

Solve for a : A = 1 2 h ( a + b )

Solve for h : V = 1 3 π r 2 h

Solve for F :   C = 5 9 ( F − 32 )

Solve for x : a x + b = c

Part C: Applications

Set up an algebraic equation then solve.

Number Problems

When 3 is subtracted from the sum of a number and 10 the result is 2. Find the number.

The sum of 3 times a number and 12 is equal to 3. Find the number.

Three times the sum of a number and 6 is equal to 5 times the number. Find the number.

Twice the sum of a number and 4 is equal to 3 times the sum of the number and 1. Find the number.

A larger integer is 1 more than 3 times another integer. If the sum of the integers is 57, find the integers.

A larger integer is 5 more than twice another integer. If the sum of the integers is 83, find the integers.

One integer is 3 less than twice another integer. Find the integers if their sum is 135.

One integer is 10 less than 4 times another integer. Find the integers if their sum is 100.

The sum of three consecutive integers is 339. Find the integers.

The sum of four consecutive integers is 130. Find the integers.

The sum of three consecutive even integers is 174. Find the integers.

The sum of four consecutive even integers is 116. Find the integers.

The sum of three consecutive odd integers is 81. Find the integers.

The sum of four consecutive odd integers is 176. Find the integers.

Geometry Problems

The length of a rectangle is 5 centimeters less than twice its width. If the perimeter is 134 centimeters, find the length and width.

The length of a rectangle is 4 centimeters more than 3 times its width. If the perimeter is 64 centimeters, find the length and width.

The width of a rectangle is one-half that of its length. If the perimeter measures 36 inches, find the dimensions of the rectangle.

The width of a rectangle is 4 inches less than its length. If the perimeter measures 72 inches, find the dimensions of the rectangle.

The perimeter of a square is 48 inches. Find the length of each side.

The perimeter of an equilateral triangle is 96 inches. Find the length of each side.

The circumference of a circle measures 80 π units. Find the radius.

The circumference of a circle measures 25 centimeters. Find the radius rounded off to the nearest hundredth.

Simple Interest Problems

For how many years must $1,000 be invested at 5 1 2 % to earn $165 in simple interest?

For how many years must $20,000 be invested at 6 1 4 % to earn $3,125 in simple interest?

At what annual interest rate must $6500 be invested for 2 years to yield $1,040 in simple interest?

At what annual interest rate must $5,750 be invested for 1 year to yield $333.50 in simple interest?

If the simple interest earned for 5 years was $1,860 and the annual interest rate was 6%, what was the principal?

If the simple interest earned for 2 years was $543.75 and the annual interest rate was 3 3 4 % , what was the principal?

How many years will it take $600 to double earning simple interest at a 5% annual rate? (Hint: To double, the investment must earn $600 in simple interest.)

How many years will it take $10,000 to double earning simple interest at a 5% annual rate? (Hint: To double, the investment must earn $10,000 in simple interest.)

Jim invested $4,200 in two accounts. One account earns 3% simple interest and the other earns 6%. If the interest after 1 year was $159, how much did he invest in each account?

Jane has her $6,500 savings invested in two accounts. She has part of it in a CD at 5% annual interest and the rest in a savings account that earns 4% annual interest. If the simple interest earned from both accounts is $303 for the year, then how much does she have in each account?

Jose put last year’s bonus of $8,400 into two accounts. He invested part in a CD with 2.5% annual interest and the rest in a money market fund with 1.5% annual interest. His total interest for the year was $198. How much did he invest in each account?

Mary invested her total savings of $3,300 in two accounts. Her mutual fund account earned 6.2% last year and her CD earned 2.4%. If her total interest for the year was $124.80, how much was in each account?

Alice invests money into two accounts, one with 3% annual interest and another with 5% annual interest. She invests 3 times as much in the higher yielding account as she does in the lower yielding account. If her total interest for the year is $126, how much did she invest in each account?

James invested an inheritance in two separate banks. One bank offered 5 1 2 % annual interest rate and the other 6 1 4 %. He invested twice as much in the higher yielding bank account than he did in the other. If his total simple interest for 1 year was $5,760, then what was the amount of his inheritance?

Uniform Motion Problems

If it takes Jim 1 1 4 hours to drive the 40 miles to work, then what is Jim’s average speed?

It took Jill 3 1 2 hours to drive the 189 miles home from college. What was her average speed?

At what speed should Jim drive if he wishes to travel 176 miles in 2 3 4 hours?

James and Martin were able to drive the 1,140 miles from Los Angeles to Seattle. If the total trip took 19 hours, then what was their average speed?

Part D: Discussion Board

What is regarded as the main business of algebra? Explain.

What is the origin of the word algebra ?

Create an identity or contradiction of your own and share it on the discussion board. Provide a solution and explain how you found it.

Post something you found particularly useful or interesting in this section. Explain why.

Conduct a web search for “solving linear equations.” Share a link to website or video tutorial that you think is helpful.

x = b + c a

  • w = P − 2 l 2
  • a = 2 A h − b
  • F = 9 5 C + 32

112, 113, 114

Width: 24 centimeters; length: 43 centimeters

Width: 6 inches; length: 12 inches

He invested $3,100 at 3% and $1,100 at 6%.

Jose invested $7,200 in the CD and $1,200 in the money market fund.

Alice invested $700 at 3% and $2,100 at 5%.

32 miles per hour

64 miles per hour

Answer may vary

Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers. 

Solution: Then the other number = x + 9 Let the number be x.  Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9)  ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides)  ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?  Solution:   Let the common ratio be x.  Let the common ratio be x.  Their difference = 48 According to the question,  7x - 3x = 48  ⇒ 4x = 48  ⇒ x = 48/4  ⇒ x = 12 Therefore, 7x = 7 × 12 = 84           3x = 3 × 12 = 36  Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.  Solution: Let the breadth of the rectangle be x,  Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72  ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x                       = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages. 

Solution: Let Ron’s present age be x.  Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4.  According to the question;  Ron will be twice as old as Aaron.  Therefore, x + 4 = 2(x - 5 + 4)  ⇒ x + 4 = 2(x - 1)  ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts.  Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x  ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x  ⇒ x = 30/2  ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40  The two parts are 15 and 25. 

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages.  Solution: Let Robert’s age be x years.  Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question,  4x + 5 = 3(x + 5)  ⇒ 4x + 5 = 3x + 15  ⇒ 4x - 3x = 15 - 5  ⇒ x = 10 ⇒ 4x = 4 × 10 = 40  Robert’s present age is 10 years and that of his father’s age = 40 years.  

7. The sum of two consecutive multiples of 5 is 55. Find these multiples.  Solution: Let the first multiple of 5 be x.  Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2  ⇒ x = 25  Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.  

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles.  Solution: Let the angle be x.  Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51  Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.  Solution: The table cost $ 40 more than the chair.  Let us assume the cost of the chair to be x.  Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x)  Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165. 

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?  Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2  According to the question,  3/5 ᵗʰ of the number is 4 more than 1/2 of the number.  ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.  

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

●   Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations

●   Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

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GCSE Maths

Linear Equations

Here we will learn how to solve linear and simple equations, including equations with one unknown, equations with an unknown on both sides, equations with brackets and equations with fractions.

There are also solving linear equations worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is a linear equation?

A linear equation is an equation that contains variables that have an exponent (are raised to a power) that is no higher than one. 

All equations have an equals sign, which means that everything on the left-hand side of the = is exactly the same as everything on the right-hand side. 

A linear equation will make a straight line graph and have a general form of ax + by + c = 0 .

This is often written as y = ax + b .

For example,

What are linear equations?

What are linear equations?

How to solve linear equations and simple equations

We solve a linear equation by combining like terms and simplifying.

There are five main types of linear and simple equations:

a Solve linear equations with one unknown

b Solve linear equations with an unknown on both sides

c Solve linear equations with brackets

d Solve linear equations with fractions

e Solve simple equations with powers (exponents) and roots

In order to solve a linear equation or a simple equation we need to work out the value of the unknown variable by doing the opposite of what the operation tells us to do.

Top tip: Leave the variable alone for as long as possible and deal with everything else first.

  • If the equation has an addition , to ‘undo it’ we need to use subtraction
  • If the equation has a subtraction , to ‘undo it’ we need to use addition
  • If the equation has a multiplication , to ‘undo it’ we need to use division
  • If the equation has a division , to ‘undo it’ we need to use multiplication

We can check that our answer is correct by substituting it back into the original equation.

What are the 5 main types of linear and simple equations?

What are the 5 main types of linear and simple equations?

Linear equations worksheet

Get your free linear equations worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on solving equations

Linear equations  is part of our series of lessons to support revision on  solving equations . You may find it helpful to start with the main solving equations lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

  • Solving equations
  • Quadratic equations
  • Equations with fractions
  • Forming and solving equations
  • Iteration maths

a) Solve linear equations with one unknown

In order to solve  linear equations with one unknown:

  • Rearrange the equation so the unknown variable is on its own on one side 
  • Work out the unknown variable by doing the opposite of what it says

Example with two steps

we need to:

1 Rearrange the equation so the unknown variable (x) is on its own on one side

Here the opposite of +6 is −6 .

2 Work out what the unknown variable (x) is by doing the opposite of what it says.

so we divide by 3 .

We can check that our solution is correct by substituting it into the original equation

Fully worked out answer:

b) Solve linear equations with an unknown on both sides

In order to solve  linear equations with an unknown on both sides:

  • Rearrange the equation so the unknown variables are on the same side
  • Rearrange the equation so the unknown variable is on its own on one side
  • Work out what the unknown variable (x) is by doing the opposite of what it says

Example with an unknown on both sides

1 Rearrange the equation so the unknown variables (x) are on the same side. (Top tip: eliminate the smallest variable).

Here 2x is smaller that 5x , so eliminate it by -2x .

2 Rearrange the equation so the unknown variable (x) is on its own on one side (here we are subtracting 6 from both sides of the equation). 

3 Work out what the unknown variable (x) is by doing the opposite of what it says.

c) Solve linear equations with brackets

In order to solve  linear equations with brackets:

Expand the brackets

  • Solve the equation by rearranging to get the unknown variable by itself on one side and then doing the opposite of what it says.

Example with brackets

1 Expand the brackets

2 Solve the equation by rearranging to get the unknown variable (x) by itself on one side and then doing the opposite of what it says.

Here the opposite of −8 is +8 .

so we divide by 4 .

We can check that our solution is correct by substituting it into the original equation.

Top Tip: We could have first divided both sides of the equation by 4 for this question as 4 is a factor of 12.

d) Solve linear equations with fractions

In order to solve linear equations with fractions:

  • Multiply each fraction by the denominator on the other side of the = to get rid of the fractions; don’t forget to include brackets.
  • Rearrange the equation so the unknown variables are on the same side.
  • Solve the equation by rearranging to get the unknown variable by itself on one side and then do the opposite of what it says.

Example with fractions

1 Multiply each fraction by the denominator on the other side of the = to get rid of the fractions; don’t forget to include brackets.

2 Expand the bracket

3 Rearrange the equation so the unknown variables (x) are on the same side. (Top tip: eliminate the smallest variable).

8x is smaller than 10x so eliminate it by -8x from both sides of the equation.

4 Solve the equation by rearranging to get the unknown variable (x) by itself on one side and then do the opposite of what it says.

e) Solve simple equations with powers and roots

In order to solve simple equations with powers and roots:

Example with a power

1 Rearrange to get the unknown variable (x 2 ) by itself on one side.

2 Work out the unknown variable by doing the opposite of what it says.

The opposite of squaring is to take the square root, so do this to both sides.

(Remember square roots have a positive and a negative answer)

Example with a square root

1 Rearrange to get the unknown variable (√x) by itself on one side.

so we divide by 5 .

2 Work out the unknown variable by doing the opposite of what it says. 

The opposite of taking the square root is squaring, so do this to both sides.

Common misconceptions

  • We must do the opposite of what the operation tells us to do

means 3 × x = 12 . So to solve the equation for x we have to do the opposite.

The opposite of × 3 is ÷ 3 .

means x ÷ 2 = 4 . So to solve the equation for x we have to do the opposite. The opposite of ÷ 2 is × 2

  • We must do the same thing to both sides of the equal sign

Equals means that both sides of the = are exactly the same.

If we −2 on the left of the equal sign, we have to −2 on the right of the equal sign

  • When cross multiplying we must use brackets to multiply every term in the numerator

We only multiply the numerator, not the denominator.

  • When expanding brackets we need to multiply out each term

Be careful when multiplying the coefficients

  • Remember a polynomial is an algebraic expression that consists of two or more algebraic terms

Practice linear equations questions

GCSE Quiz False

Subtract 8 from both sides

Divide both sides by 2

Add 3 to both sides

Divide both sides by 5

Subtract x from both sides

Add 7 to both sides

Subtract 4x from both sides

Divide both sides by 3

3(x-3)=6(x+5)

Subtract 3x from both sides

Subtract 30 from both sides

\frac{x+3}{2}=\frac{x+4}{3}

Multiply both sides by 6 and simplify

From both sides, subtract 2x an d subtract 9

\frac{3x+1}{2}+\frac{2x-2}{4} 

\frac{3x+1}{2}+\frac{2x-2}{4}

Multiply both sides by 4 and simplify

From both sides, subtract 2x and subtract 2

 16{x}^2=64

Divide both sides by 16

Square root both sides

5\sqrt{3x}=30

\sqrt{3x}=6

Square both sides

Linear equations GCSE questions

3 x^{2} = 27  

            (1 mark)

4(3 – x) = 32  

            (2 marks)

12 – 4x = 32  or  3 – x = 8 

\frac{2x + 3}{5}=\frac{x – 5}{2}  

            (3 marks)

2(2x + 3) = 5(x – 5)   

4x + 6 = 5x – 25 

Learning checklist

You have now learned how to:

  • Use algebraic methods to solve linear equations.

The next lessons are:

  • Factorising
  • Simultaneous equations
  • Rearranging equations

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4.2 Solving linear equations

4.2 solving linear equations (ema34).

The simplest equation to solve is a linear equation. A linear equation is an equation where the highest exponent of the variable is \(\text{1}\). The following are examples of linear equations:

Solving an equation means finding the value of the variable that makes the equation true. For example, to solve the simple equation \(x + 1 = 1\), we need to determine the value of \(x\) that will make the left hand side equal to the right hand side. The solution is \(x = 0\).

The solution, also called the root of an equation, is the value of the variable that satisfies the equation. For linear equations, there is at most one solution for the equation.

To solve equations we use algebraic methods that include expanding expressions, grouping terms, and factorising.

For example:

Check the answer by substituting \(x=-\frac{1}{2}\).

Therefore \(x=-\frac{1}{2}\)

The following video gives an introduction to solving linear equations.

Video: 2F9B

Method for solving linear equations (EMA35)

The general steps for solving linear equations are:

Expand all brackets.

Rearrange the terms so that all terms containing the variable are on one side of the equation and all constant terms are on the other side.

Group like terms together and simplify.

Factorise if necessary.

Find the solution and write down the answer.

Check the answer by substituting the solution back into the original equation.

An equation must always be balanced, whatever you do to the left-hand side, you must also do to the right-hand side.

Worked example 1: Solving linear equations

Solve for \(x\):

Expand the brackets and simplify

Divide both sides by 10, check the answer by substituting the solution back into the original equation.

Since both sides are equal, the answer is correct.

Worked example 2: Solving linear equations

Multiply both sides of the equation by \(\left(3x + 1\right)\).

Division by \(\text{0}\) is undefined so there must be a restriction: \(\left(x\ne -\frac{1}{3}\right)\).

Divide both sides by \(-\text{7}\)

Worked example 3: solving linear equations.

Solve for \(a\): \[\frac{2a - 3}{3} - 3a = \frac{a}{3}\]

Multiply the equation by the common denominator \(\text{3}\) and simplify

Rearrange the terms and simplify, divide both sides by \(-\text{8}\).

Solve the following equations (assume all denominators are non-zero):

\(2y - 3 = 7\)

\(2c = c -8\)

\(3 = 1 - 2c\)

\begin{align*} 3 &= 1 - 2c\\ 2c &= 1 - (3)\\ 2c & = -2\\ c & = \frac{-2}{2}\\ & = -1 \end{align*}

\(4b+5 = -7\)

\begin{align*} 4b +5 &= -7\\ 4b &= -7 - (5)\\ 4b & = -12\\ b & = \frac{-12}{4}\\ & = -3 \end{align*}

\(-3y = 0\)

\(16y + 4 = -10\)

\(12y + 0 = 144\)

\(7 + 5y = 62\)

\(55 = 5x + \frac{3}{4}\)

\(5x = 2x + 45\)

\(23x - 12 = 6 + 3x\)

\(12 - 6x + 34x = 2x - 24 - 64\)

\(6x + 3x = 4 - 5(2x - 3)\)

\(18 - 2p = p + 9\)

\(\dfrac{4}{p} = \dfrac{16}{24}\)

\(-(-16 - p) = 13p - 1\)

\(3f - 10 = 10\)

\(3f + 16 = 4f - 10\)

\(10f + 5 = -2f -3f + 80\)

\(8(f - 4) = 5(f - 4)\)

\(6 = 6(f + 7) + 5f\)

\(-7x = 8(1 - x)\)

\(5 - \dfrac{7}{b} = \dfrac{2(b + 4)}{b}\)

\(\dfrac{x + 2}{4} - \dfrac{x - 6}{3} = \dfrac{1}{2}\)

Note that \(a \neq - -3\)

Note \(b \neq -5\)

\(3 - \dfrac{y - 2}{4} = 4\)

\(\text{1,5}x + \text{3,125} = \text{1,25}x\)

\(\frac{1}{3}P + \frac{1}{2}P - 10 = 0\)

\(1\frac{1}{4}(x - 1) - 1\frac{1}{2}(3x + 2) = 0\)

\(\dfrac{5}{2a} + \dfrac{1}{6a} - \dfrac{3}{a} = 2\)

Linear Equations Questions

The linear equations questions and answers will assist students to understand the concepts better. Linear Equation is a topic that is covered in basically every class. The NCERT guidelines will be followed for preparing the questions. Linear Equations are used in mathematics as well as in everyday life. So, the basics of this concept must be grasped. For students of all levels, the problems here will include both the basics and more challenging problems. As a result, students will be able to use it to solve problems involving linear equations. Learn more about Linear Equations by clicking here .

Here, we’ll go through a variety of linear equations problems with solutions, based on various concepts.

Linear Equations Questions with Solutions

1. Write the statement as an equation: A number increased by 8 equals 15.

Given statement: A number increased by 8 equals 15.

Let the number be “x”.

So, x is increased by 8 means x + 8.

Hence, x increased by 8 equals 15 means x + 8 = 15, which is the equation for the given statement.

2. Write the statement for the given equation: 2x = 18.

Given equation: 2x = 18.

The statement for the given equation is “Twice the number x equals 18”.

3. Solve the equation: x + 3 = -2

Given equation: x + 3 = -2.

Now, keep the variables on one side and constants on the other side. Hence, the equation becomes,

Hence, the value of x is -5.

4. Verify that x = 4 is the root of the equation 3x/2 = 6.

To verify whether the given root is the solution of the given equation, substitute x = 4 in the equation 3x/2 = 6.

⇒ (3(4))/2 = 6

⇒ (12/2) = 6

Hence, x = 4 is the root of the equation 3x/2 = 6.

5. If 5 is added to twice a number, the result is 29. Determine the number.

The equation for the given statement is 5+2x = 29.

To find the number “x”, we have to solve the equation.

⇒ 2x = 29 – 5

Hence, the required number is 12.

6. If x = 2, then 2x – 5 = 7. Check whether the statement is true or false.

Given equation: 2x – 5 = 7

= 2(2) – 5

= 4 – 5 = -1

Hence, the given statement is false.

7. The sum of two consecutive numbers is 11. Find the numbers.

Let the number be x.

Hence, the two consecutive numbers are x and x+1.

According to the given statement, the equation becomes

⇒ x + x + 1 = 11

⇒ 2x + 1 = 11

⇒ x = 10/2 = 5

If x = 5, then x + 1 = 5 + 1 = 6

Hence, the two numbers are 5 and 6.

8. Express the equation x = 3y in the form of ax+by+c = 0 and find the values of a, b and c.

Given equation: x = 3y

We know that the standard form of linear equation in two variables is ax+by+c = 0 …(1)

Now, rearranging the given equation, we get

⇒ x – 3y = 0

This can be written as

⇒ 1(x) + (-3)y + (0)c = 0 …(2)

On comparing equation (1) and (2), we get

⇒ a = 1, b = -3 and c = 0.

9. Find three solutions for the equation 2x + y = 7.

To find the solutions for the equation 2x + y = 7, substitute different values for x.

When x = 0,

⇒ 2(0) + y = 7

Therefore, the solution is (0, 7).

When x = 1,

⇒ 2(1) + y = 7

⇒ y = 7 – 2

Hence, the solution is (1, 5).

When x = 2,

⇒ 2(2) + y = 7

⇒ 4 + y = 7

Hence, the solution is (2, 3).

Therefore, the three solutions are (0, 7), (1, 5) and (2, 3).

10. Solve the following equations using the substitution method:

3x + 4y = 10 and 2x – 2y = 2

3x + 4y = 10 …(1)

2x – 2y = 2 …(2)

Equation (2) can be written as:

2(x – y) = 2

x – y = 1

x = 1+y …(3)

Now, substitute (3) in (1), we get

3 (1+y) + 4y = 10

3 + 3y + 4y = 10

7y = 10 – 3

Hence, y = 1.

Now, substitute y = 1 in (3), we get

Hence, x = 2 and y = 1 are the solutions of the given equations.

Practice Questions

  • Write the statement as an equation: Twice a number subtracted from 19 is 11.
  • The sum of the two numbers is 30 and their ratio is 2: 3. Find the numbers.
  • If the point (3, 4) lies on the graph of equation 3y = ax + 7, determine the value of a.
  • Solve the equations using the elimination method: (x/2)+(2y/3) = -1 and x – (y/3) = 3.

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COMMENTS

  1. What Are Some Real Life Examples of Linear Equations?

    Real-life examples of linear equations include distance and rate problems, pricing problems, calculating dimensions and mixing different percentages of solutions. Linear equations are used in the form of mixing problems, where different per...

  2. Resources to Help You Solve Math Equations

    Whether you love math or suffer through every single problem, there are plenty of resources to help you solve math equations. Skip the tutor and log on to load these awesome websites for a fantastic free equation solver or simply to find an...

  3. What Is the History of Linear Equations?

    Linear algebra originated as the study of linear equations and the relationship between a number of variables. Linear algebra specifically studies the solution of simultaneous linear equations.

  4. Linear Equations (Practice Problems)

    Section 2.2 : Linear Equations · 4x−7(2−x)=3x+2 4 x − 7 ( 2 − x ) = 3 x + 2 Solution · 2(w+3)−10=6(32−3w) 2 ( w + 3 ) − 10 = 6 ( 32 − 3 w ) Solution · 4−2

  5. Examples on Solving Linear Equations

    Examples on Solving Linear Equations: ; 1. Solve: (2x + 5)/(x + 4) = 1. Solution: (2x + 5)/(x + 4) = 1 ; ⇒ 2x - x = 4 - 5 (Transferring positive x to the left

  6. Solving Linear Equations

    Next, isolate the variable using the equality property of multiplication or division. Checking the solution in the following examples is left to the reader.

  7. Word Problems on Linear Equations

    Word Problems on Linear Equations · Then the other number = x + 9. Let the number be x. · 2.The difference between the two numbers is 48. · 3. The length of a

  8. SOLVING LINEAR EQUATIONS

    This type of equation is called a contradiction. All other linear equations which have only one solution are called conditional. Examples: A. Check: (

  9. Solve Linear Equations Practice

    1. Solve for x: 3x - 12 = 0. basket1aa · 2. Solve for m: 2(m + 6) = 48. purplegirl1 · 3. Solve for x: 3(2x - 1) - 10 = 8 + 5x. basket2 · 4. Solve for x: 8x + 9 -

  10. Linear Equations

    There are also solving linear equations worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you're still

  11. Linear(Simple) Equations: Problems with Solutions

    Solve the equation z - 5 = 6. ... Solve the equation 5 - t = 0. ... Solve the linear equation 3-a=2a. ... Find c

  12. 4.2 Solving linear equations

    Solving an equation means finding the value of the variable that makes the equation true. For example, to solve the simple equation x+1=1

  13. SOLVING LINEAR EQUATIONS.pdf

    common denominator and proceed as usual. Model Problems: Solve:.

  14. Linear Equations Questions with Solutions

    1. Write the statement as an equation: A number increased by 8 equals 15. · 2. Write the statement for the given equation: 2x = 18. · 3. Solve the equation: x + 3