5.6 Solving Equations Using Both the Addition and Multiplication Properties of Equality

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1 5.6 Solving Equations Using Both the Addition and Multiplication Properties of Equality Now that we have studied the Addition Property of Equality and the Multiplication Property of Equality, we can solve equations that require the use of both properties. In this chapter, we have been solving linear equations in one variable, which are equations that contain only one variable, and that variable is raised only to the first power. The following are examples of linear equations in one variable that we will study in this section. x + = = = x 6x 8x + 2 = x (x + 2) = 7 STEPS TO SOLVING LINEAR EQUATIONS IN ONE VARIABLE. If the expression on either side of the equation has like terms, combine those like terms. 2. Move all variable terms to one side of the equation and all constant terms to the other side of the equation using the Addition Property of Equality.. Get the variable by itself on one side of the equation using the Multiplication Property of Equality.. Check the answer in the original equation. If that answer produces a true statement, then the answer is the solution. 0

2 Example : Solve the equation x + =. Then check the solution. Let s try to do this by thinking about what the equation says. times what number plus equals? First, ask yourself what number plus equals? We know that 8 plus equals. So x must equal 8. So, times what number equals 8? The number is 2. This was a little more difficult to work through. Now, we will learn how to solve this equation algebraically. To solve the equation means to determine the value of the variable that makes the equation a true statement. To do this, we want to get the variable on one side of the equation by itself; we call this isolating the variable. We will do this using the steps for solving equations. x x - - x 8 x 8 x 2. The expressions on the right and left of the equal sign are already simplified. Skip to step Subtract from both sides.. Divide both sides of the equation by. Check: Substitute 2 for x in the original equation. ( 2) 8 This is a true statement. Therefore, x = 2 is the solution of x. Looking at Example, we can see that each equation in the solving process looks a little different from the preceding one. What is interesting and useful is that each of the equations says the same thing about x: each one says that x is 2. The last equation, of course, is the easiest to read, which is why our goal is to end up with x isolated on one side of the equation. Practice : Solve the equation 7x + 2 = 2. Answer: x = 05

3 Example 2: Solve the equation = -. Then check the solution x 2. The expressions on the right and left of the equal sign are already simplified. Skip to step Add to both sides of the equation.. Divide both sides of the equation by 2. Check: Substitute - 2 for x in the original equation Substitute for x. 2 Write as a fraction multiplication problem. Divide out a factor of 2 in the numerator and denominator. This is a true statement. Therefore, x = - 2 is the solution of = -. Practice 2: Solve the equation 5x 9 = 2. Answer: x =

4 Example : Solve the equation 2 2 x. x 2 2 x x 0 x (0) The check is left to you. x 0 0 x 0 x. The expressions on the right and left of the equal sign are already simplified. Skip to step Subtract 2 from both sides of the equation.. Multiply both sides of the equation by, the reciprocal of the coefficient of x. Divide out common factors in the numerator and denominator. Practice : Solve the equation 2 x 2 0. Answer: x = 8 Example : Solve the equation The expressions on the right and left of the equal sign are already simplified. Skip to step Add.7 to both sides of the equation x Divide both sides of the equation by 0.2. Simplify The check is left to you. Practice : Solve the equation 0.5x Answer: x =

5 Example 5: Solve the equation x. Then check the solution x x.5 0.5x.5 0.5x x Check: Substitute 7 for x in the original equation x ( 7) The expressions on the right and left of the equal sign are already simplified. Skip to step Add.5 to both sides of the equation.. Divide both sides of the equation by 0.5. Simplify 2 2 This is a true statement. Therefore, x = 7 is the solution of x. Practice 5: Solve the equation Answer: x = Example 6: Solve the equation 5xx 5 9. Then check the solution. 5xx 5 9 Combine like terms on the left side of the equal sign. Combine like terms on the right side of the equal sign. Subtract from both sides of the equation x Divide both sides of the equation by 2, the coefficient of x. Check: Substitute - for x in the original equation. 5x x 5 9 5( -) ( -) This is a true statement. Therefore, x = - is the solution of 5xx

6 Practice 6: Solve the equation5x 0. Answer: x = Example 7: Solve the equation 2 6x. Then check the solution x x 22 To combine like terms on the left side of the equal sign the fractions must have common denominators. Multiply the first fraction s numerator and denominator by 2 in order to make a common denominator of in both fractions x 5 Add the numerators and keep the same 2 6x denominator x Subtract 2 from both sides of the equation x 5 2 6x Write 2 as the fraction 2 in order to subtract the fractions. Multiply the first fraction s numerator and denominator by in order to make a common denominator of in both fractions x Add the numerators and keep the same 6x denominator. ( 6 x) Multiply both sides of the equation by the 6 6 reciprocal of the coefficient of x. On the left side, divide out a in the numerator x and denominator. On the right side, use the 2 6 inverse property of multiplication to simplify. x 8 09

7 Check: Substitute 8 for x in the original equation. 2 6x This is a true statement. Therefore, x is the solution of 2 6x. 8 2 Practice 7: Solve the equation 5. 8 Answer: Example 8: Solve the equation5( x) x 5 8. Then check the solution. 5( x) x x - x x Distribute.. Simplify the left and right sides of the equation by combining like terms. 2. Subtract 5 from both sides of the equation.. Divide both sides of the equation by 2. 0

8 Check: Substitute - for x in the original equation. 5( x) x 5 8 5( ( ) ) ( ) 5 8 5( ) ( ) ( 2) This is a true statement. Therefore, x = - is the solution of 5( x) x 5 8. Practice 8: Solve the equation 2( x) x 7. Answer: x = - Watch All:

9 5.6 Solving Equations using the Addition and Multiplication Property Exercises Solve each equation.. 5x = x = = 2y. x = = y x 7. 2 x x a + = 5. 9 = 7y = x x. 6x 9x + 2 = x = x t x 7x 9. ( + ) x = (x + 5) 6 = 2 2

10 5.6 Addtion and Multiplication Property Exercises Answers. x = 8 2. x = = y. x = = y 6. x = x or 7 8. x x.6 0. a =. -6 = y 2. x = -. x =. x = 0 5. x = x 0 7. t 8. x 2 9. x = x =

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