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Solve Multistep Equations Honors Math – Grade 8 Get Ready for the Lesson In 2003, about 46.9 million U.S. households had dial-up Internet service and about 26 million had broadband service. During the next five years, it was projected that the number of dial-up users would decrease an average of 3 million per year and the number of broadband users would increase an average of 8 million per year. The following expressions represent the number of dial-up and broadband users x years after 2003. Dial-up Users: 46.9 – 3x Broadband: 26 + 8x To find the time at which Dial-up users and broadband users are the same number, set the equations equal to each other. 46.9 3x 26 8x To solve an equation the variable on both sides, use the Addition & Subtraction Properties of Equality to write an equation with all the variables on one side and the constants on the other. Solve each equation. First, move one of the variables. Since 10p is greater than 8p, leave 10p on the left and move 8p. Subtract 8p from both sides of the equation. Add 2 to both sides of the equation. Divide both sides of the equation by 2. Answers may be left in fractional form or written as a decimal. Don’t forget to check your solution! 2 10 p 8 p 1 8 p 8 p 2 2 p 1 2 2 2p 1 2 2 1 p 2 Solve each equation. First, move one of the variables. Since 7w is greater than 3w, leave 7w on the right and move 39. Subtract 3p from both sides of the equation. Divide both sides of the equation by 2. Answers may be left in fractional form or written as a decimal. Don’t forget to check your solution! 3w 2 7w 3w 3w 2 4w 4 4 1 w 2 Solve each equation. To eliminate the denominators, think what is the LCD of ½ and ¼ ? The LCD is 4. Multiply both sides of the equation by 4. Distribute the 4 on each side of the equation. Since 2s is greater than 1s, move 1s to the other side of the equation. Subtract s from both sides of the equation. Subtract 4 from both sides of the equation. Don’t forget to check your solution! s 1 4 1 4 s 6 2 4 2s 4 s 24 s s s 4 24 4 4 s 28 Solve each equation. If an equation contains grouping symbols, first use the Distributive Property to remove the grouping symbols. 4 2r ? 4 8 ? 1 1 49r ? 70 ? 7 7 Since 8r is greater than 7r, move 7r to the other side of the equation. Subtract 7r from both sides of the equation. Add 32 to both sides of the equation. 1 4(2r 8) (49r 70) 7 8r 32 7r 10 7r 7r r 32 10 32 32 r 42 Solve each equation. If an equation contains grouping symbols, first use the Distributive Property to remove the grouping symbols. 7 n ? 7 1 ? 23 ? 2 n ? Rewrite plus a negative! Since 7n is greater than 2n, move 2n to the other side of the equation. Add 2n to both sides of the equation. Add 7 to both sides of the equation. 7(n 1) 2(3 n) 7 n 7 6 (2n) 7n 7 6 2n 2n 2n 9n 7 6 7 7 9n 1 9 9 1 n 9 Solve each equation. If an equation contains grouping symbols, first use the Distributive Property to remove the grouping symbols. 3 6 ? 3 2s ? Since 8s is greater than 6s, move 6s to the other side of the equation. Add 6s to both sides of the equation. Add 10 to both sides of the equation. 8s 10 3(6 2s) 8s 10 18 6s 6s 14s 10 18 10 10 14s 28 14 14 s2 6s Solve each equation. Before collecting unknown terms, you may first have to group like terms. Since 7v is greater than 3v, leave 7v on the left and move 3v. Add 3v to both sides of the equation. Subtract 9 from both sides of the equation. Divide both sides of the equation by 10. Answers may be left in fractional form or written as a decimal. REDUCE! Don’t forget to check your solution! 6v 9 v 4 3v 7v 9 4 3v 3v 3v 10v 9 4 9 9 10v 5 1 10 10 v 2 Solve each equation. If an equation contains grouping symbols, first use the Distributive Property to remove the grouping symbols. 5 m ? 5 7 ? 2m 5 5(m 7) 3m 2m 5 5m 35 3m 2m 5 2m 35 2m 2m Group like terms. Subtract 2m from both sides of the equation. 5 35 Since 5 = -35 is a false statement, the original equation is a contradiction. The solution set is the null set. Solve each equation. If an equation contains grouping symbols, first use the Distributive Property to remove the grouping symbols. 3 r ? 3 1 ? Group like terms. 3(r 1) 5 3r 2 3r 3 5 3r 2 3r 2 3r 2 Since the expressions on each side of the equation are the same, this equation illustrates the Reflexive Property of Equality. This equation is called an identity. It is true for all values of r. The solution set is all real numbers!