2-4 Solving Equations with the Variable on Each Side. Solve each equation. Check your solution. 13x + 2 = 4x Check: Check: Page 1

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1 Solve each equation. Check your solution. 13x + 2 = 4x + 38 Page 1

2 6(n + 4) = 18 7 = (b + 5) Page 2

3 6(n + 4) = 18 7 = (b + 5) 5 + 2(n + 1) = 2n Page 3

4 5 + 2(n + 1) = 2n Since, this equation has no solutions. To check, substitute any number for n. 7 4(2 + r) 3r = r Since, this equation has no solutions. To check, substitute any number for r. 14v + 6 = 2(5 + 7v) 4 Page 4

5 14v + 6 = 2(5 + 7v) 4 Since the expressions on each side are the same, the equation is an identity. So, all numbers will work in this equation. To check the solution, substitute any number for v. 5h 7 = 5(h 2) + 3 Since the expressions on each side are the same, the equation is an identity. So, all numbers will work in this equation. To check the solution, substitute any number for h. MULTIPLE CHOICE Find the value of x so that the figures have the same perimeter. A B Page 5

6 MULTIPLE CHOICE Find the value of x so that the figures have the same perimeter. A B C D The following is an expression for the perimeter of the triangle. Let a = 3x + 4, b = 5x + 1, and c = 2x + 5. The following is an expression for the perimeter of the rectangle.let and w = 2x. Set the expressions equal to each other and solve for x. Page 6

7 When x = 4, both figures have a perimeter of 50 units. Choice A is correct. Solve each equation. Check your solution. 7c + 12 = 4c m 13 = 8m + 27 Page 7

8 2m 13 = 8m x 4 = 2x t = 8t 14 Page 8

9 6 + 3t = 8t 14 Page 9

10 8 = 4(r + 4) 6(n + 5) = 66 Page 10

11 8 = 4(r + 4) 6(n + 5) = 66 5(g + 8) 7 = 103 Page 11

12 5(g + 8) 7 = 103 3(3m 2) = 2(3m + 3) Page 12

13 3(3m 2) = 2(3m + 3) 6(3a + 1) 30 = 3(2a 4) GEOMETRY Page 13

14 6(3a + 1) 30 = 3(2a 4) GEOMETRY Find the value of x so the rectangles have the same area. The following is an expression for the area of the left rectangle. The following is an expression for the area of the right rectangle. Set the expressions equal to each other and solve for x. Page 14

15 GEOMETRY Find the value of x so the rectangles have the same area. The following is an expression for the area of the left rectangle. The following is an expression for the area of the right rectangle. Set the expressions equal to each other and solve for x. Find the areas to check the solution. When x NUMBER THEORY Four times the lesser of two consecutive even integers is 12 less than twice the greater number. Find the integers. Let x be the first even integer and x + 2 be the next consecutive even integer. Page 15

16 When x NUMBER THEORY Four times the lesser of two consecutive even integers is 12 less than twice the greater number. Find the integers. Let x be the first even integer and x + 2 be the next consecutive even integer. Substitute 4 into each expression to find the even numbers. x= 4 x + 2 = or 2 So, the consecutive even integers are 4 and 2. CCSS SENSE-MAKING Two times the least of three consecutive odd integers exceeds three times the greatest by 15. What are the integers? Let x = the least odd integer. Then x + 2 = the next greater odd integer, and x + 4 = the greatest of the three integers. Substitute 27 into each expression to find the odd numbers. x = 27 x + 2 = or 25 x + 4 = or 23 So, the consecutive odd integers are 27, 25, and 23. Solve each equation. Check your solution. 2x = 2(x 3) Page 16

17 x = 27 x + 2 = or 25 x + 4 = or 23 So, the consecutive odd integers are 27, 25, and 23. Solve each equation. Check your solution. 2x = 2(x 3) Since, this equation has no solutions. To check, substitute any number for x. Since, this equation has no solutions. To check, substitute any number for h. 5(3 q) + 4 = 5q 11 Since the expressions on each side are the same, the equation is an identity. So, all numbers will work in this equation. To check, substitute any number for q. Page 17

18 5(3 q) + 4 = 5q 11 Since the expressions on each side are the same, the equation is an identity. So, all numbers will work in this equation. To check, substitute any number for q. Since the expressions on each side are the same, the equation is an identity. So, all numbers will work in this equation. To check, substitute any number for r. Page 18

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22 6.78j 5.2 = 4.33j t 25.2 = 3.8t Page 22

23 14.2t 25.2 = 3.8t k 4.3 = 12.6k [2p 4(p + 5)] = 25 Page 23

24 5[2p 4(p + 5)] = 25 NUMBER THEORY Three times the lesser of two consecutive even integers is 6 less than six times the greater number. Find the integers. Let x be the first consecutive even integer and x + 2 be the next consecutive even integer. Substitute 2 into each expression to find the even numbers. x= 2 x + 2 = or 0 So, the consecutive even integers are 2 and 0. MONEY Chris has saved twice the number of quarters that Nora saved plus 6. The number of quarters Chris saved is also five times the difference of the number of quarters and 3 that Nora has saved. Write and solve an equation to find the number of quarters they each have saved. Page 24 Let q be the number of quarters Nora saved.

25 Substitute 2 into each expression to find the even numbers. x= 2 x + 2 = or 0 So, the consecutive even integers are 2 and 0. MONEY Chris has saved twice the number of quarters that Nora saved plus 6. The number of quarters Chris saved is also five times the difference of the number of quarters and 3 that Nora has saved. Write and solve an equation to find the number of quarters they each have saved. Let q be the number of quarters Nora saved. Substitute 7 for q to find the number of quarters Chris saved. So, Nora saved 7 quarters and Chris saved 20 quarters. DVD A company that replicates DVDs spends $1500 per day in building overhead plus $0.80 per DVD in supplies and labor. If the DVDs sell for $1.59 per disk, how many DVDs must the company sell each day before it makes a profit? Let d be the number of DVDs. So, the company must sell more than 1,899 DVDs per day to make a profit. MOBILE PHONES The table shows the number of mobile phone subscribers for two states for a recent year. How long will Let t be the number of years. Let x be the number of years. Page 25

26 So, the company must sell more than 1,899 DVDs per day to make a profit. MOBILE PHONES The table shows the number of mobile phone subscribers for two states for a recent year. How long will Let t be the number of years. Let x be the number of years x x 325x 325x 292x 33x = = = = = x x x x 292x 77 So it will take about 2.3 years or 2 years 4 months to be the same MULTIPLE REPRESENTATIONS In this problem, you will explore 2x + 4 = x 2. a. GRAPHICAL Make a table of values with five points for y = 2x + 4 and y = x 2. Graph the points from the tables. b. ALGEBRAIC Solve 2x + 4 = x 2. c. VERBAL Explain how the solution you found in part b is related to the intersection point of the graphs in part a. a. b. Page 26 c. Draw the lines y = 2x + 4 and y = x 2. Then find the intersection of the lines.

27 325x 292x 33x = = x x So it willequations take about with 2.3 years or 2 yearson 4 months to be the same 2-4 Solving the Variable Each Side MULTIPLE REPRESENTATIONS In this problem, you will explore 2x + 4 = x 2. a. GRAPHICAL Make a table of values with five points for y = 2x + 4 and y = x 2. Graph the points from the tables. b. ALGEBRAIC Solve 2x + 4 = x 2. c. VERBAL Explain how the solution you found in part b is related to the intersection point of the graphs in part a. a. b. c. Draw the lines y = 2x + 4 and y = x 2. Then find the intersection of the lines. The lines intersect when x = 2. This is the x-coordinate for the point of intersection on the graph. REASONING Solve 5x + 2 = ax 1 for x. Assume that a Page 27

28 The lines intersect when x = 2. This is the x-coordinate for the point of intersection on the graph. REASONING Solve 5x + 2 = ax 1 for x. Assume that a CHALLENGE Write an equation with the variable on each side of the equals sign, at least one fractional coefficient, and a solution of 6. Discuss the steps you used. First, chose a fractional coefficient. Then, chose a coefficient for the variable on the other side of the equation. 2x Plugging in 6 on both sides. To balance, 1 must be added to the left and 2 must be subtracted. OPEN ENDED Create an equation with at least two grouping symbols for which there is no solution. Page 28

29 OPEN ENDED Create an equation with at least two grouping symbols for which there is no solution.. CCSS CRITIQUE Determine whether each solution is correct. If the solution is not correct, describe the error and give the correct solution. a. b. c. a. This is incorrect. The 2 must be distributed over both g and 5. The correct answer is g = 6. b. c. This is incorrect. To eliminate 6z on the left side of the equals sign, 6z must be added to each side of the equation. esolutions Manual - Powered by Cognero Page 29

30 . CCSS CRITIQUE Determine whether each solution is correct. If the solution is not correct, describe the error and give the correct solution. a. b. c. a. This is incorrect. The 2 must be distributed over both g and 5. The correct answer is g = 6. b. c. This is incorrect. To eliminate 6z on the left side of the equals sign, 6z must be added to each side of the equation. The correct answer is z = 1. CHALLENGE Find the value esolutions Manual - Powered by Cognero a. k(3x 2) = 4 6x b. 15y 10 + k = 2(k y 1) y of k for which each equation is an identity. Page 30

31 The correct answer is z = 1. CHALLENGE Find the value of k for which each equation is an identity. a. k(3x 2) = 4 6x b. 15y 10 + k = 2(k y 1) y a. An equation that is true for every value of the variable is an identity. Find the value k always 4 6x k(3x The equation is an identity when k = 2. b. Find the value k y 10 + k is always 2(k y 1) y The equation is an identity when k = 8. WRITING IN MATH Compare and contrast solving equations with variables on both sides of the equation to solving one-step or multi-step equations with a variable on one side of the equation. If the equation has variables on both sides of the equation, you must first add or subtract one of the terms from both sides of the equation so that the variable is left on only one side of the equation. Then, solving the equations uses the same steps. x = 7x Consider the multi-step problem 8 + 5x = 2. 2 with variables on both side. Page 31

32 The equation is an identity when k = 8. WRITING IN MATH Compare and contrast solving equations with variables on both sides of the equation to solving one-step or multi-step equations with a variable on one side of the equation. If the equation has variables on both sides of the equation, you must first add or subtract one of the terms from both sides of the equation so that the variable is left on only one side of the equation. Then, solving the equations uses the same steps. x = 7x 2 with variables on both side. Consider the multi-step problem 8 + 5x = 2. After step 4, the steps for solving an equation with variable on both size are the same as a multi-step equation. A hang glider 25 meters above the ground starts to descend at a constant rate of 2 meters per second. Which equation shows the height h after t seconds of descent? A h = 25t + 2t B h = 25t + 2 C h = 2t + 25 D h = 2t + 25 The hang glider is starting at 25 feet. He is falling at 2 feet per second. Falling suggests a negative. So, the equation that could be used to determine the hang glider s height after t seconds is h = 2t +25. Choice D is correct. GEOMETRY Two rectangular walls each with a length of 12 feet and a width of 23 feet need to be painted. It costs $0.08 per square foot for paint. How much will it cost to paint the walls? F $22.08 G $23.04 H $34.50 J $44.16 Page 32

33 D h = 2t + 25 The hangequations glider is starting at 25 feet. Heon is Each fallingside at 2 feet per second. Falling suggests a negative. So, the equation 2-4 Solving with the Variable that could be used to determine the hang glider s height after t seconds is h = 2t +25. Choice D is correct. GEOMETRY Two rectangular walls each with a length of 12 feet and a width of 23 feet need to be painted. It costs $0.08 per square foot for paint. How much will it cost to paint the walls? F $22.08 G $23.04 H $34.50 J $44.16 Since there are two walls, the total area that needs covered is 2 the total cost, multiply the total area by the cost per square foot, which is 552 paint the walls. Choice J is correct. 0.08, or So it costs $44.16 to SHORT RESPONSE Maddie works at Game Exchange. They are having a sale as shown. Her employee discount is 15%. If sales tax is 7.25%, how much does she spend for a total of 4 video games? First, find Maddie s discount = 3 Now, subtract her discount from the original cost = 17 Maddie will pay $17 per video game. For every two she buys, she gets one free. Since she is getting 4, she will buy two, get 1 free, and then buy another. So, she will only pay for 3 video games = 51 Her total cost for the games is $ Now find the sales tax = Rounded to the nearest cent, the sales tax is $3.70. To find her total cost, add the tax to the cost of the games. So, Maddie paid , or $ Solve A B D 10 Page 33

34 3 17 = 51 Her total cost for the games is $ Now find the sales tax = RoundedEquations to the nearest sales tax $3.70. To find her total cost, add the tax to the cost of the games. So, 2-4 Solving withcent, the the Variable oniseach Side Maddie paid , or $ Solve A B D 10 Choice A is correct. Solve each equation. Check your solution. 5n + 6 = 4 1 = 7 + 3c Page 34

35 Choice A is correct. Solve each equation. Check your solution. 5n + 6 = 4 1 = 7 + 3c Page 35

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38 WORLD RECORDS In 1998, Winchell s House of Donuts in Pasadena, California, made the world s largest donut. It weighed 5000 pounds and had a circumference of feet. What was the donut s diameter to the nearest tenth? (Hint: C = d) We are given the circumference, so we can use this value and the formula for circumference to find the diameter. The diameter of the donut is about 95.0 feet. ZOO At a zoo, the cost of admission is posted on the sign. Find the cost of admission for two adults and two children. Zoo admission for two adults and two children will cost $34. Find the value of n. Then name the property used in each step. 25n = 25 Page 38

39 The diameter of the donut is about 95.0 feet. ZOO At a zoo, the cost of admission is posted on the sign. Find the cost of admission for two adults and two children. Zoo admission for two adults and two children will cost $34. Find the value of n. Then name the property used in each step. 25n = 25 Since the product of 25 and 1 is 25, the property used is the Multiplicative Identity. n 1=2 Since the product of 2 and 1 is 2, the property used is the Multiplicative Identity. 12 n = 12 6 When 6 is substituted back into the equation, each side is equal to 72. Since 72 = 72, the property used is the Reflexive Property. n+0=, the property used is the Additive Identity. Page 39

40 When 6 Equations is substituted back into the equation, eachside side is equal to 72. Since 72 = 72, the property used is the 2-4 Solving with the Variable on Each Reflexive Property. n+0=, the property used is the Additive Identity. Since the product of 4 and (10 8)(7) = 2(n) Since the first quantity of (10 8) on the left is equal to the first quantity 2 on the right and the second quantity 7 on the left is equal to the second quantity n or 7 on the right, the property used is the Transitive Property. Translate each sentence into an equation. Twice a number t decreased by eight equals seventy. Rewrite the verbal sentence so it is easier to translate. Twice a number t decreased by eight equals seventy is the same as 2 times t minus 8 equals times The equation is 2t t t minus 8 8 equals = = 70. Five times the sum of m and k is the same as seven times k. 5 5 times the sum of m and k The equation is 5(m + k) = 7k. is the same as 7 = 7 times k k Page 40

41 2 2 times minus t t 8 8 equals = Solving Equations the Variable on Each Side The equation is 2t with 8 = 70. Five times the sum of m and k is the same as seven times k. 5 times 5 the sum of m and k is the same as 7 = 7 times k k The equation is 5(m + k) = 7k. Half of p is the same as p minus 3. Half of The equation is p is the same as p p = p p =p minus Evaluate each expression. 9 ( 14) 10 + (20) 10 + (20) = (14) 5(14) = 70 5) 55 5 = 11 25( 5) 25( 5) = 125 Page 41

42 5) 55 5 = 11 25( 5) 25( 5) = 125 Page 42