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hendricks_intermediate_algebra_1e_ch1_3

Section 2.1 Solving Linear Equations 57 Procedure: General Strategy for Solving Linear Equations Step 1: Write an equivalent equation that doesn’t contain fractions or decimals, if necessary. a. Clear fractions from the equation by multiplying each side by the LCD of the fractions in the equation. b. Clear decimals from the equation by multiplying by an appropriate power of 10. Step 2: Clear any parentheses by applying the distributive property. Step 3: Apply the addition property of equality to isolate variable terms on one side of the equation and constant terms on the other side. Step 4: Apply the multiplication property of equality to obtain a coefficient of 1 on the variable. Step 5: Check by substituting the proposed solution in the original equation. Objective 4 Examples Solve each linear equation. 4a. 4a - 9 + 5a = 2a - 7 + 3 4b. -2(3x + 4) = 6 - 4(x - 2) 4c. 2m 3 + 3 2 = 2m 4d. 0.05(x - 2) + 0.10x = 3.65 Solutions 4a. 4a - 9 + 5a = 2a - 7 + 3 9a - 9 = 2a - 4 9a - 9 - 2a = 2a - 4 - 2a 7a - 9=-4 7a - 9 + 9=-4 + 9 7a = 5 7a 7 = 5 7 a = 5 7 Check: Combine like terms on each side of the equation. Subtract 2a from each side. Simplify. Add 9 to each side. Simplify. 4a - 9 + 5a = 2a - 7 + 3 Begin with the original equation. 4¢ 5 7 ≤ - 9 + 5¢ 5 7 ≤ = 2¢ 5 7 ≤ - 7 + 3 Replace a with 5 7 . 20 7 - 9 + 25 7 = 10 7 - 4 Simplify each product. 20 7 - 63 7 + 25 7 = 10 7 - 28 7 Convert each number to a fraction with an LCD of 7. - 18 7 =- 18 7 Simplify each side. The value 5 7 makes the equation true, so the solution set is e 5 7 f . Divide each side by 7. Simplify.


hendricks_intermediate_algebra_1e_ch1_3
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