1.4 Solving Multi-Step Equations

To isolate the variable, perform the inverse or opposite of every operation in the equation on both sides of the equation. Do inverse operations in the reverse order of operations.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Example 2 Continued

Distribute 4.

Distribute before solving.

4m + 48 = –36

4m = –84

–48 –48 Subtract 48 from both sides.

Divide both sides by 4.=4m –84 4 4

m = –21

Solve 4(m + 12) = –36

Method 2

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Distribute 3.

Method 2

Distribute before solving.

6 – 9p = 42

–9p = 36

–6 –6 Subtract 6 from both sides.

Divide both sides by –9.=–9p 36–9 –9

p = –4

Check It Out! Example 2a Continued

Solve 3(2 – 3p) = 42 .

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Distribute 3.

Distribute before solving.

–15 + 12r = –9

12r = 6

+15 +15Add 15 to both sides.

Divide both sides by 12.=12r 6

12 12

Check It Out! Example 2b Continued

r =

Solve –3(5 – 4r) = –9.

Method 2

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

If there are variables on both sides of the equation, (1) simplify each side. (2) collect all variable terms on one side and all constants terms on the other side. (3) isolate the variables as you did in the previous problems.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Example 3: Solving Equations with Variables on Both Sides

Simplify each side by combining like terms.–11k + 25 = –6k – 10

Collect variables on the right side.

Add.

Collect constants on the left side.

Isolate the variable.

+11k +11k

25 = 5k – 10

35 = 5k

5 5

7 = k

+10 + 10

Solve 3k– 14k + 25 = 2 – 6k – 12.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Check It Out! Example 3

Solve 3(w + 7) – 5w = w + 12.

Simplify each side by combining like terms.–2w + 21 = w + 12

Collect variables on the right side.

Add.

Collect constants on the left side.

Isolate the variable.

+2w +2w

21 = 3w + 12

9 = 3w

3 3

3 = w

–12 –12

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

You have solved equations that have a single solution. Equations may also have infinitely many solutions or no solution.

An equation that is true for all values of the variable, such as x = x, is an identity. An equation that has no solutions, such as 3 = 5, is a contradiction because there are no values that make it true.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve 3v – 9 – 4v = –(5 + v).

Example 4A: Identifying Identities and Contractions

3v – 9 – 4v = –(5 + v)

Simplify.–9 – v = –5 – v

+ v + v

–9 ≠ –5 x Contradiction

The equation has no solution..

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve 2(x – 6) = –5x – 12 + 7x.

Example 4B: Identifying Identities and Contractions

2(x – 6) = –5x – 12 + 7x

Simplify.2x – 12 = 2x – 12

–2x –2x

–12 = –12 Identity

The solutions set is all real numbers.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve 5(x – 6) = 3x – 18 + 2x.

The equation has no solution.

Check It Out! Example 4a

5(x – 6) = 3x – 18 + 2x

Simplify.5x – 30 = 5x – 18

–5x –5x

–30 ≠ –18 x Contradiction

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve 3(2 –3x) = –7x – 2(x –3).

3(2 –3x) = –7x – 2(x –3)

Simplify.6 – 9x = –9x + 6

+ 9x +9x

6 = 6 Identity

The solutions set is all real numbers.

Check It Out! Example 4b