2.5 Solve Equations with variables on Both SidesEssential Question: How do you solve multi-step

equations?Warm-up: Solve the equation.1. 2m – 6 + 4m = 122. 6a – 5 (a – 1) = 11.3. A charter bus company charges $11.25 per ticket plus a handling charge of $.50 per ticket, and a $15 fee for booking the bus. If a group pays $297 a charter a bus, how many tickets did they buy?

CommonCoreCC.9-12.A.REI.3

9-12-14

Solve linear equations andinequalities in one variable, including equations with coefficients represented byletters.

EXAMPLE 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

EXAMPLE 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

EXAMPLE 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

GUIDED PRACTICE for Examples 1 and 2

3ANSWER

1. 24 – 3m = 5m

Solve the equation. Check your solution.

GUIDED PRACTICE for Examples 1 and 2

2. 20 + c = 4c – 7

ANSWER 9

Solve the equation. Check your solution.

GUIDED PRACTICE for Examples 1 and 2

3. 9 – 3k = 17k – 2k

Solve the equation. Check your solution.

ANSWER –8

GUIDED PRACTICE for Examples 1 and 2

4. 5z – 2 = 2(3z – 4)

Solve the equation. Check your solution.

ANSWER 6

GUIDED PRACTICE for Examples 1 and 2

5. 3 – 4a = 5(a – 3)

Solve the equation. Check your solution.

ANSWER 2

GUIDED PRACTICE for Examples 1 and 2

8y – 6 =23 (6y + 15)6.

ANSWER 4

Solve the equation. Check your solution.

Solve 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?

Solve 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) )

Solve a real-world problem

EXAMPLE 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.

Solve 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

GUIDED 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

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.

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

EXAMPLE 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

GUIDED PRACTICE for Example 4

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

Solve the equation, if possible.

ANSWER

no solution

GUIDED PRACTICE for Example 4

9. 7w + 1 = 8w + 1

ANSWER

0

Solve the equation, if possible.

GUIDED PRACTICE for Example 4

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

ANSWER

identity

Solve the equation, if possible.

Summary: Steps for Solving Linear Equations

Step 1: Use the distributive property to remove any grouping symbols.

Step 2: Simplify the expression on each side of the equation.

Step 3: Use properties of equality to collect the variable terms on one side of the equation and the constant terms on the other side of the equation.

Step 4: Use properties of equality to solve for the variable.

Step 5: Check your solution in the original equation.

Classwork/Homework

•2.5 Exercises•2-54 even

•Pages 107-109