2.5 Solving equations with variables on both sides

• You will solve equations with variables on both sides.

• Essential Question: How do you solve equations with variables on both sides?

Warm-Up ExercisesEXAMPLE 1 Solve an equation with variables on both sides

7 – 8x = 4x – 17

7 – 8x + 8x = 4x – 17 + 8x

7 = 12x – 17

24 = 12x

Write original equation.

Add 8x to each side.

Simplify each side.

Add 17 to each side.

Divide each side by 12.

ANSWER

The solution is 2. Check by substituting 2 for x in the original equation.

Solve 7 – 8x = 4x – 17.

2 = x

Warm-Up ExercisesEXAMPLE 1 Solve an equation with variables on both sides

Write original equation.

Substitute 2 for x.

Simplify left side.

Simplify right side. Solution checks.

–9 = 4(2) – 17?

7 – 8(2) = 4(2) – 17?

7 – 8x = 4x – 17 CHECK

–9 = –9

Warm-Up ExercisesEXAMPLE 2 Solve an equation with grouping symbols

14

(16x + 60)9x – 5 =

9x – 5 = 4x + 15

5x – 5 = 15

5x = 20

x = 4

Write original equation.

Distributive property

Subtract 4x from each side.

Add 5 to each side.

Divide each side by 5.

9x – 5 =14 (16x + 60).Solve

Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2

3ANSWER

1. 24 – 3m = 5m

Solve the equation. Check your solution.

Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2

2. 20 + c = 4c – 7

ANSWER 9

Solve the equation. Check your solution.

Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2

3. 9 – 3k = 17k – 2k

Solve the equation. Check your solution.

ANSWER –8

Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2

4. 5z – 2 = 2(3z – 4)

Solve the equation. Check your solution.

ANSWER 6

Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2

5. 3 – 4a = 5(a – 3)

Solve the equation. Check your solution.

ANSWER 2

Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2

8y – 6 =23 (6y + 15)6.

ANSWER 4

Solve the equation. Check your solution.

Warm-Up ExercisesSolve a real-world problemEXAMPLE 3

CAR SALES

A car dealership sold 78 new cars and 67 used cars this year. The number of new cars sold by the dealership has been increasing by 6 cars each year. The number of used cars sold by the dealership has been decreasing by 4 cars each year. If these trends continue, in how many years will the number of new cars sold be twice the number of used cars sold?

Warm-Up ExercisesSolve a real-world problemEXAMPLE 3

SOLUTION

Let x represent the number of years from now. So, 6x represents the increase in the number of new cars sold over x years and –4x represents the decrease in the number of used cars sold over x years. Write a verbal model.

6778 + 6x = 2 ( + (– 4 x) )

Warm-Up ExercisesSolve a real-world problemEXAMPLE 3

78 + 6x = 2(67 – 4x)

78 + 6x = 134 – 8x

78 + 14x = 134

14x = 56

x = 4

Write equation.

Distributive property

Add 8x to each side.

Subtract 78 from each side.

Divide each side by 14.

ANSWER

The number of new cars sold will be twice the number of used cars sold in 4 years.

Warm-Up ExercisesSolve a real-world problemEXAMPLE 3

CHECK You can use a table to check your answer.

YEAR 0 1 2 3 4

Used car sold 67 63 59 55 51

New car sold 78 84 90 96 102

Warm-Up ExercisesGUIDED PRACTICE for Example 3

7.

WHAT IF? In Example 3, suppose the car dealership sold 50 new cars this year instead of 78. In how many years will the number of new cars sold be twice the number of used cars sold?

ANSWER

6 yr

Warm-Up Exercises

SOLUTION

EXAMPLE 4 Identify the number of solutions of an equation

Solve the equation, if possible.

a. 3x = 3(x + 4) b. 2x + 10 = 2(x + 5)

a. 3x = 3(x + 4) Original equation

3x = 3x + 12 Distributive property

The equation 3x = 3x + 12 is not true because the number 3x cannot be equal to 12 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

Warm-Up Exercises

ANSWER

The statement 0 = 12 is not true, so the equation hasno solution.

Simplify.

3x – 3x = 3x + 12 – 3x Subtract 3x from each side.

0 = 12

EXAMPLE 4 Identify the number of solutions of an equation

Warm-Up ExercisesEXAMPLE 1

b. 2x + 10 = 2(x + 5) Original equation

2x + 10 = 2x + 10 Distributive property

ANSWER

Notice that the statement 2x + 10 = 2x + 10 is true for all values of x. So, the equation is an identity, and the solution is all real numbers.

EXAMPLE 4 Identify the number of solutions of an equation

Warm-Up ExercisesGUIDED PRACTICE for Example 4

8. 9z + 12 = 9(z + 3)

Solve the equation, if possible.

ANSWER

no solution

Warm-Up ExercisesGUIDED PRACTICE for Example 4

9. 7w + 1 = 8w + 1

ANSWER

0

Solve the equation, if possible.

Warm-Up ExercisesGUIDED PRACTICE for Example 4

10. 3(2a + 2) = 2(3a + 3)

ANSWER

identity

Solve the equation, if possible.

Warm-Up ExercisesDaily Homework Quiz

Solve the equation, if possible.

3(3x + 6) = 9(x + 2)1.

7(h – 4) = 2h + 172.

ANSWER The equation is an identity.

ANSWER 9

8 – 2w = 6w – 83.

ANSWER 2

Warm-Up ExercisesDaily Homework Quiz

4g + 3 = 2(2g + 3)4.

ANSWER The equation has no solution.

ANSWER 5 h

Bryson is looking for a repair service for general household maintenance. One service charges $75 to join the service and $30 per hour. Another service charge $45 per hour. After how many hours of service is the total cost for the two services the same?

5.

• You will solve equations with variables on both sides.

• Essential Question: How do you solve equations with variables on both sides?

•To solve equations with variables on both sides, collect the variable terms on one side and the constant terms on the other.• Some equations, called identities, are true for all values of the variable. Other equations have no solutions.

To solve equations with variableson both sides, first simplify theexpressions on each side of theequation by using the distributiveproperty to remove grouping symbolsand then combining like terms.Next, use properties of equality tocollect variable terms on one sideof the equation and constants onthe other. Then solve the equationby isolating the variable.