Section 2.4 More on Solving Linear Equations 129 2b. 0.04x + 0.05(500 - x) = 23 1000.04x + 0.05(500 - x) = 100(23) Multiply each side by 100. Distribute and simplify. Simplify each product. Apply the distributive property. Combine like terms. Subtract 2500 from each side. Simplify. Divide each side by -1. Simplify. 100(0.04x) + 100(0.05)(500 - x) = 2300 4x + 5(500 - x) = 2300 4x + 2500 - 5x = 2300 -x + 2500 = 2300 -x + 2500 - 2500 = 2300 - 2500 -x=-200 -x = -1 -200 -1 x = 200 Check: 0.04x + 0.05(500 - x) = 23 Original equation Replace x with 200. Simplify. Multiply. Add. 0.04(200) + 0.05(500 - 200) = 23 8 + 0.05(300) = 23 8 + 15 = 23 23 = 23 Since x = 200 makes the equation true, the solution set is {200}. Student Check 2 Solve each equation by first clearing the decimals. Check each answer. a. x + 0.25x = 90 b. 0.05x + 0.07(800 - x) = 54 Linear Equations with No Solution All of the linear equations we have encountered so far have had one solution. When solving these equations, we were able to isolate the variable on one side of the equation and the constant on the other side. These types of equations are called conditional equations. ������������������������ A conditional equation is an equation that is true for some values of the variable and not true for other values of the variable. We will now turn our attention to a special type of linear equation, one that has no solution. Consider the equation x = x + 1. The left and right sides of this equation will never be equal since the right side of the equation is always one more than the left side of the equation as shown in the table. x x = x + 1 -2 -2 = -2 + 1 -2 = -1 False -1 -1 = -1 + 1 -1 = 0 False 0 0 = 0 + 1 0 = 1 False 1 1 = 1 + 1 1 = 2 False 2 2 = 2 + 1 2 = 3 False So, x = x + 1 is an example of an equation with no solution. It is called a contradiction. Objective 3 ▶ Solve linear equations with no solution.
hendricks_beginning_algebra_1e_ch1_3
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