3.5 Solving Equations Involving Integers II
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1 208 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA 3.5 Solving Equations Involving Integers II We return to solving equations involving integers, only this time the equations will be a bit more advanced, requiring the use of the distributive property and skill at combining like terms. Let s begin. Solve for x: 6x 5x = 22 EXAMPLE 1. Solve for x: 7x 11x = 12. Solution. Combine like terms. 7x 11x = 12 4x = 12 Combine like terms: 7x 11x = 4x. To undo the effect of multiplying by 4, divide both sides of the last equation by 4. 4x 4 = 12 4 Divide both sides by 4. x = 3 Simplify: 12/( 4) = 3. Check. Substitute 3 for x in the original equation. 7x 11x = 12 7( 3) 11( 3) = 12 Substitute 3 for x = 12 On the left, multiply first. 12 = 12 On the left, add. Answer: x = 2 Because the last line of the check is a true statement, 3 is a solution of the original equation. Solve for x: 11 = 3x (1 x) EXAMPLE 2. Solve for x: 12 = 5x (4+x). Solution. To take the negative of a sum, negate each term in the sum (change each term to its opposite). Thus, (4+x) = 4 x. 12 = 5x (4+x) 12 = 5x 4 x (4+x) = 4 x. 12 = 4x 4 Combine like terms: 5x x = 4x. To undo the effect of subtracting 4, add 4 to both sides of the last equation = 4x 4+4 Add 4 to both sides. 16 = 4x Simplify both sides.
2 3.5. SOLVING EQUATIONS INVOLVING INTEGERS II 209 To undo the effect of multiplying by 4, divide both sides of the last equation by = 4x 4 Divide both sides by 4. 4 = x Simplify: 16/4 = 4. Check. Substitute 4 for x in the original equation. 12 = 5x (4+x) 12 = 5(4) (4+4) Substitute 4 for x. 12 = 20 8 On the right, 5(4) = 20 and evaluate 12 = 12 Simplify. parentheses: 4+4 = 8. Because the last line of the check is a true statement, 4 is a solution of the original equation. Answer: x = 3 Variables on Both Sides Variables can occur on both sides of the equation. Goal. Isolate the terms containing the variable you are solving for on one side of the equation. EXAMPLE 3. Solve for x: 5x = 3x 18. Solve for x: Solution. To isolate the variables on one side of the equation, subtract 3x from both sides of the equation and simplify. 4x 3 = x 5x = 3x 18 5x 3x = 3x 18 3x 2x = 18 Subtract 3x from both sides. Combine like terms: 5x 3x = 2x and 3x 3x = 0. Note that the variable is now isolated on the left-hand side of the equation. To undo the effect of multiplying by 2, divide both sides of the last equation by 2. 2x 2 = 18 2 Divide both sides by 2. x = 9 Simplify: 18/2 = 9.
3 210 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA Check. Substitute 9 for x in the original equation. 5x = 3x 18 5( 9) = 3( 9) 18 Substitute 9 for x. 45 = Multiply first on both sides. 45 = 45 Subtract on the right: = 45. Answer: x = 1 Because the last line of the check is a true statement, 9 is a solution of the original equation. Solve for x: 7x = 18+9x EXAMPLE 4. Solve for x: 5x = 3+6x. Solution. To isolate the variables on one side of the equation, subtract 6x from both sides of the equation and simplify. 5x = 3+6x 5x 6x = 3+6x 6x x = 3 Subtract 6x from both sides. Combine like terms: 5x 6x = x and 6x 6x = 0. Note that the variable is now isolated on the left-hand side of the equation. There are a couple of wayswe can finish this solution. Remember, x is the same as ( 1)x, so we could undo the effects of multiplying by 1 by dividing both sides of the equation by 1. Multiplying both sides of the equation by 1 will work equally well. But perhaps the easiest way to proceed is to simply negate both sides of the equation. ( x) = 3 Negate both sides. x = 3 Simplify: ( x) = x. Check. Substitute 3 for x in the original equation. 5x = 3+6x 5( 3) = 3+6( 3) Substitute 3 for x. 15 = 3 18 Multiply first on both sides. 15 = 15 Subtract on the right: 3 18 = 15. Answer: x = 9 Because the last line of the check is a true statement, 3 is a solution of the original equation.
4 3.5. SOLVING EQUATIONS INVOLVING INTEGERS II 211 Dealing with x. If your equation has the form x = c, where c is some integer, note that this is equivalent to the equation ( 1)x = c. Therefore, dividing both sides by 1 will produce a solution for x. Multiplying both sides by 1 works equally well. However, perhaps the easiest thing to do is negate each side, producing ( x) = c, which is equivalent to x = c. EXAMPLE 5. Solve for x: 6x 5 = 12x+19. Solve for x: Solution. To isolate the variables on one side of the equation, subtract 12x from both sides of the equation and simplify. 6x 5 = 12x+19 6x 5 12x = 12x+19 12x Subtract 12x from both sides. 6x 5 = 19 Combine like terms: 6x 12x = 6x and 12x 12x = 0. Note that the variable is now isolated on the left-hand side of the equation. Next, to undo subtracting 5, add 5 to both sides of the equation. 6x 5+5 = 19+5 Add 5 to both sides. 6x = 24 Simplify: 5+5 = 0 and 19+5 = 24. Finally, to undo multiplying by 6, divide both sides of the equation by 6. 6x 6 = 24 6 Divide both sides by 6. x = 4 Simplify: 24/( 6) = 4. Check. Substitute 4 for x in the original equation. 2x+3 = 18 3x 6x 5 = 12x+19 6( 4) 5 = 12( 4)+19 Substitute 4 for x = Multiply first on both sides. 29 = 29 Add: 24 5 = 29 and = 29. Because the last line of the check is a true statement, 4 is a solution of the original equation. Answer: x = 3
5 212 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA Solve for x: 3(2x 4) 2(5 x) = 18 EXAMPLE 6. Solve for x: 2(3x+2) 3(4 x) = x+8. Solution. Usethedistributivepropertytoremoveparenthesesontheleft-hand side of the equation. 2(3x+2) 3(4 x) = x+8 6x x = x+8 9x 8 = x+8 Use the distributive property. Combine like terms: 6x+3x = 9x and 4 12 = 8. Isolatethevariablesontheleftbysubtractingxfrombothsidesoftheequation. 9x 8 x = x+8 x 8x 8 = 8 Subtract x from both sides. Combine like terms: 9x x = 8x and x x = 0. Note that the variable is now isolated on the left-hand side of the equation. Next, to undo subtracting 8, add 8 to both sides of the equation. 8x 8+8 = 8+8 Add 8 to both sides. 8x = 16 Simplify: 8+8 = 0 and 8+8 = 16. Finally, to undo multiplying by 8, divide both sides of the equation by 8. 8x 8 = 16 8 Divide both sides by 8. x = 2 Simplify: 16/8 = 2. Check. Substitute 2 for x in the original equation. 2(3x+2) 3(4 x) = x+8 2(3(2)+2) 3(4 2) = 2+8 Substitute 2 for x. 2(6+2) 3(2) = 10 2(8) 3(2) = = 10 Work parentheses on left, add on the right. Add in parentheses on left. Multiply first on left. 10 = 10 Subtract on left. Answer: x = 5 Because the last line of the check is a true statement, 2 is a solution of the original equation.
6 3.5. SOLVING EQUATIONS INVOLVING INTEGERS II 213 Exercises In Exercises 1-16, solve the equation. 1. 9x+x = x 5x = = 3x 4x 4. 6 = 5x+7x 5. 27x+51 = x+46 = = 5x+9 6x 8. 6 = x+3 4x 9. 0 = 18x = x = 28x = x x 8 9x = x+7 9x = x+85 = x 17 = 0 In Exercises 17-34, solve the equation x = 5x x = 3x x 7 = 5x 20. 3x+8 = 5x 21. 4x 3 = 5x x 2 = 9x x+5 = 3x x+9 = 4x x = 3x x = 4x x 2 = 4x 28. 6x 4 = 2x 29. 6x+8 = 2x 30. 4x 9 = 3x 31. 6x = 4x x = 6x x+2 = 6x x+6 = 2x 5 In Exercises 35-52, solve the equation (x 2) = (x 8) = x+6(x+8) = x+4(x+7) = ( 6x 1) = ( 2x 4) = ( 4x 6) = (2x+8) = (x 5) = (x+4) = 1
7 214 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA 45. 7x+2(x+9) = x+7(x 2) = ( x+8) = ( x 2) = (x 5) = (x+5) = x 2(x+5) = x 5(x 3) = 15 In Exercises 53-68, solve the equation ( 7x+5)+8 = 3( 9x 1) ( x+9)+5 = ( 5x 4) ( 2x 6) = 7(5x 1) ( 4x 8) = 9( 6x+4) (2x 9)+5 = 7( x 8) 58. 6( 4x 9)+4 = 2( 9x 8) 59. 6( 3x+4) 6 = 8(2x+2) (5x 9) 3 = 4(2x+5) ( 2x 3) = 3( x+2) 62. 2(7x+1) = 2(3x 7) 63. 5( 9x+7)+7 = ( 9x 8) 64. 7( 2x 6)+1 = 9( 2x+7) 65. 5(5x 2) = 4(8x+1) 66. 5( x 4) = ( x+8) 67. 7(9x 6) = 7(5x+7) (2x+1) = 2( 9x+8) 2 Answers
8 3.5. SOLVING EQUATIONS INVOLVING INTEGERS II
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