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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. 2-1 Solving Linear Equations and Inequalities Example 2 Continued Solve 4(m + 12) = –36 Method 2 Distribute before solving. 4m + 48 = –36 –48 –48 Distribute 4. Subtract 48 from both sides. 4m = –84 4m –84 = 4 4 m = –21 Holt Algebra 2 Divide both sides by 4. 2-1 Solving Linear Equations and Inequalities Check It Out! Example 2a Continued Solve 3(2 – 3p) = 42 . Method 2 Distribute before solving. 6 – 9p = 42 –6 –6 –9p = 36 –9p 36 = –9 –9 p = –4 Holt Algebra 2 Distribute 3. Subtract 6 from both sides. Divide both sides by –9. 2-1 Solving Linear Equations and Inequalities Check It Out! Example 2b Continued Solve –3(5 – 4r) = –9. Method 2 Distribute before solving. –15 + 12r = –9 +15 +15 Distribute 3. 12r = 6 12r 12 6 = 12 r= Holt Algebra 2 Add 15 to both sides. Divide both sides by 12. 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 Solve 3k– 14k + 25 = 2 – 6k – 12. –11k + 25 = –6k – 10 +11k +11k Collect variables on the right side. 25 = 5k – 10 +10 + 10 35 = 5k 5 5 7=k Holt Algebra 2 Simplify each side by combining like terms. Add. Collect constants on the left side. Isolate the variable. 2-1 Solving Linear Equations and Inequalities Check It Out! Example 3 Solve 3(w + 7) – 5w = w + 12. –2w + 21 = +2w w + 12 +2w Collect variables on the right side. 21 = 3w + 12 –12 –12 9 = 3w 3 3 3=w Holt Algebra 2 Simplify each side by combining like terms. Add. Collect constants on the left side. Isolate the variable. 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 Example 4A: Identifying Identities and Contractions Solve 3v – 9 – 4v = –(5 + v). 3v – 9 – 4v = –(5 + v) –9 – v = –5 – v +v Simplify. +v –9 ≠ –5 x The equation has no solution.. Holt Algebra 2 Contradiction 2-1 Solving Linear Equations and Inequalities Example 4B: Identifying Identities and Contractions Solve 2(x – 6) = –5x – 12 + 7x. 2(x – 6) = –5x – 12 + 7x 2x – 12 = 2x – 12 –2x Simplify. –2x –12 = –12 The solutions set is all real numbers. Holt Algebra 2 Identity 2-1 Solving Linear Equations and Inequalities Check It Out! Example 4a Solve 5(x – 6) = 3x – 18 + 2x. 5(x – 6) = 3x – 18 + 2x 5x – 30 = 5x – 18 –5x –5x –30 ≠ –18 The equation has no solution. Holt Algebra 2 Simplify. x Contradiction 2-1 Solving Linear Equations and Inequalities Check It Out! Example 4b Solve 3(2 –3x) = –7x – 2(x –3). 3(2 –3x) = –7x – 2(x –3) 6 – 9x = –9x + 6 + 9x Simplify. +9x 6=6 The solutions set is all real numbers. Holt Algebra 2 Identity