Section 2.1 Solving Linear Equations 53 However, -10 is not a solution of the linear equation 5x - 5 = 45 since the resulting equation is not true as shown. 5x - 5 = 45 5(-10) - 5 = 45 Replace x with -10. -50 - 5 = 45 Simplify the left side. -55 = 45 False Procedure: Determining if a Value Is a Solution of an Equation Step 1: Replace the variable with the given value. Step 2: Simplify each side of the resulting equation. Step 3: If the simplified equation is true, the value is a solution. Otherwise, it is not a solution. Objective 1 Examples Determine if the equation is a linear equation. If the equation is linear, determine if -3 is a solution of the equation. 1a. 4x - 10 = 2 1b. 3x + 7 2 =-1 1c. 2x2 + 3x = 11 Solutions 1a. The equation 4x - 10 = 2 is a linear equation because the exponent of the variable is 1. To determine if -3 is a solution, we replace the variable with -3. 4x - 10 = 2 Begin with the equation. 4(-3) - 10 = 2 Replace x with -3. -12 - 10 = 2 Multiply. -22 = 2 Simplify. Since -3 makes the equation false, it is not a solution of 4x - 10 = 2. 1b. The equation 3x + 7 2 =-1 is equivalent to 3 2 x + 7 2 =-1. It is linear because the exponent of the variable is 1. To determine if -3 is a solution, we replace the variable with -3. 3x + 7 2 =-1 Begin with the equation. 3(-3) + 7 2 =-1 Replace x with -3. -9 + 7 2 =-1 Multiply. -2 2 =-1 Add. -1=-1 Simplify. Since -3 makes the equation true, it is a solution of 3x + 7 2 =-1. 1c. The equation is not linear because the largest exponent on the variable is 2. Student Check 1 Determine if the equation is a linear equation. If the equation is linear, determine if - 1 2 is a solution of the equation. a. 8x - 3=-7 b. 6(x + 1) + 5=-2 c. 4x2 - 2x 2 = 5
hendricks_intermediate_algebra_1e_ch1_3
To see the actual publication please follow the link above