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Olympic College Topic 2 – Solving Equations Topic 2 Solving Equations Introduction: When you are given the value of a variable and an algebraic expression then you can evaluate the expression. For example, If you are told that x = 6 then the value of the expression 2x + 7 can be found by

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  1. Olympic College Topic 2 – Solving Equations Topic 2 Solving Equations Introduction: When you are given the value of a variable and an algebraic expression then you can evaluate the expression. For example, If you are told that x = 6 then the value of the expression 2x + 7 can be found by replacing the x with its value 6 and calculating the result as shown here, 2x + 7 = 2(6) + 7 = 12 + 7 = 19 Now imagine a situation where you are given the value of an expression instead, Say we are told that 5x + 12 = 32 and now we wish to find the value of the variable x. What we are doing in this situation is solving the equation 5x + 12 = 32 This section deals with the algebraic processes that can be used to solve these “Linear Equations” . Definition: A “Linear Equation” is in an equation that can be written in the form ax + b = c where a,b and c are real numbers. In other words to be a linear equation it can only contain the variable x and constant terms. For example. The following are all examples of “Linear Equations”. 2x = 12 3x + 7 4x – 8 2(3x – 7) = = 9 2x – 7 23 + 7x = There are many different types of equations and so we need to adjust the methods used to take account of each situation. The main types of equations are as follows. Type 1: Type 2: Type 3: Type 4: Type 5: Simple Add and Subtract Equations. Multiply Equations. Simple Mixed Equations. Distributive Law Equations. Complex Mixed Equations. Page | 1

  2. Olympic College Topic 2 – Solving Equations Type 1 Simple Add and Subtract Equations These equations have a general form x + a = b where a and b are real numbers. To solve these equations we use a balance technique which uses the fact that as long as you do the same operation to both sides of an equation then the resulting equation will be equivalent to the original equation. This process can be used to simplify and eventually solve the equation. Example 1: Solve the equation x + 7 = 12 Solution: x+7 = 12 x+7 – 7 12 – 7 5 = = Subtract 7 from both sides x Notice in this situation we need to remove the + 7 from the right hand side of the equation and we achieve this by subtracting 7 from both sides of the equation. We can also check the solution by substitution. Check: Using x = 5 in the algebraic expression x + 7 = 5 + 7 = 12 This check proves we have a correct solution as we got the expected value of 12. Example 2: Solve the equation x – 7 = 12 x – 7 Solution: = 12 x – 7+7 = = 12 + 7 19 Add 7 from both sides x Notice in this situation we need to remove the – 7 from the right hand side of the equation and we achieve this by adding 7 to both sides of the equation. We can also check the solution by substitution. Check: Using x = 19 in the algebraic expression x – 7 = 19 – 7 = 12 This check proves we have a correct solution as we got the expected value of 12. Solve the following equations:– Exercise 1A: x – 4=6 k + 5 =18 x – 7=0 t – 5=9 1. 5. x + 5 = 12 d – 9=2 2. 6. 3. 7. 4. 8. x+7=9 y + 9 = 12 11. x– 4 = 67 12. r – 31 =56 9. e+1=4 10. p + 9 = 19 13. y + 13 = – 30 17. u – 67 = –93 21. k + 7 = 3 14. d – 7 = – 20 18. m – 5 = 6 22. k + 17 = 16 15. y + 5 = – 8 19. s – 5 = 27 23. k – 9 = 7 16. h – 3 = – 15 20. b + 5 = 65 24. x + 2 = 2 Page | 2

  3. Olympic College Topic 2 – Solving Equations Type 2 Multiply Equations. These equations have a general form ax = b where a and b are real numbers. To solve these equations we use a balance technique which uses the fact that as long as you do the same operation to both sides of an equation then you can use this method to solve the equation. In this case we typically divide both sides by a Example 1: Solve the equation 4y = 24 Solution: 4y = = 24 Divide both sides by 4 y = 6 Notice in this situation we need to remove the 4 from the left hand side of the equation and we achieve this by dividing both sides of the equation by 4. Check: Using y= 6 in the algebraic expression 4y = 4(6) = 24 This check proves we have a correct solution as we got the expected value of 24. Example 2: Solve the equation – 3x = 15 – 3x x Solution: = = = 15 – 5 Divide both sides by – 3 Notice in this situation we need to remove the – 3 from the left hand side of the equation and we achieve this by dividing both sides of the equation by – 3 . Check: Using x = – 5 in the algebraic expression – 3x = – 3(– 5) = 15 This check proves we have a correct solution as we got the expected value of 15. Example 3: Solve the equation x = 7 Solution: x = 7 2( x) = 2(7) Multiply both sides by 2 x = 14 Notice in this situation we need to remove the achieve this by multiply both sides by 2. from the left hand side of the equation and we Check: Using x = 14 in the algebraic expression x = (14) = 7 Page | 3

  4. Olympic College Topic 2 – Solving Equations This check proves we have a correct solution as we got the expected value of 7. Exercise 2A: 1. Solve the following simple equations. (a) 2x = (c) 5x = (e) 2x = (g) 8b = (i) – 5d = (k) – 3x = 6 60 – 6 – 24 25 – 9 (b) 4x = (d) 8x = (f) 3x = (h) – 2c = (j) – 6x= (l) – 5x= 16 16 – 15 6 – 12 0 (m) ½ x = 5 (n) = 10 (o) = 10 (p) ¼ x = 5 = – 10 (q) (r) = 10 Page | 4

  5. Olympic College Topic 2 – Solving Equations Type 3 Simple Mixed Equations These equations have a general form ax + b = c where a, b and c are real numbers. To solve these equations we use the usual balance technique of subtracting b from both sides and then dividing both sides by a. Example 1: Solve the equation 4x + 9 = 21 Solution: 4x + 9 = 21 4x + 9 – 9 21 – 9 = Subtract 9 from both sides 4x = = 12 Divide both sides by 4 x = 3 Check: Using x= 3 in the algebraic expression 4x + 9 = 4(3) + 9 = 21 This check proves we have a correct solution as we got the expected value of 21. Example 2: Solve the equation 2 = 2x + 14 Solution: 2 = 2x + 14 2 – 14 – 12 2x + 14 – 14 2x = = Subtract 14 from both sides = – 6 = Divide both sides by 2 x Check: Using x= – 6 in the algebraic expression 2x + 14 = 2(– 6) + 14 = 2 This check proves we have a correct solution as we got the expected value of 2. Solve the following equations:– Exercise 3A: 3x – 1 = 20 6a + 3 = 27 1. 5. 2x + 1 = 15 4y + 4 = 12 2. 6. 3. 7. 2x + 4 = 10 3p + 2 =14 4. 8. 3x + 7 = 22 3x – 4 = 11 6x – 2 = 10 10. 3r – 2 = 10 11. 7t – 4 = 17 12. 10y – 2 = 18 9. 14. 9r – 4 = 41 18. 7b + 15 = 78 22. 7 = 5x + 2 26. 7b + 5 = 75 30. 70 = – 5x + 15 15. 3c – 8 = 19 19. – 8c – 7 = 65 23 = – 3e + 3 27. 2x – 1 = 5 31. 10 = 4x + 2 13. 7x + 5 = 26 17. – 6d – 19 = 11 21. 71 = 6f + 17 25. 6d – 19 = 1 29. 71 = – 6b + 69 16. 9t + 8 = 53 20. 7b + 15 = 71 24. 45 = 7x + 3 28. 7b + 15 = 15 32. 45 = 2x + 3 Page | 5

  6. Olympic College Topic 2 – Solving Equations Type 4: Distributive Law Equations. These equations will contain parenthesis and we must first remove the parenthesis and then we can solve the resulting equation. Example 1: Solve the equation 4(x + 2) = 36 Solution: 4(x + 2) = 36 4x + 8 = 36 Using the Distributive Law 4x = = 28 Subtract 8 from both sides Divide both sides by 4 x = 7 Check: Using x= 7 in the algebraic expression 4(x + 2) = 4(7 + 2) = 4(9) = 36 This check proves we have a correct solution as we got the expected value of 36. Example 2: Solve the equation 6(2x + 2) = 24 Solution: 6(2x + 2) = 24 12x + 12 12x = = = = 24 12 1 Using the Distributive Law Subtract 12 from both sides Divide both sides by 12 x Check: Using x= 1 in the algebraic expression 6(2x + 2)= 6(2(1) + 2)= 6(2 + 2) = 6(4) = 24 This check proves we have a correct solution as we got the expected value of 2. Example 3: Solve the equation 5(x – 2) = – 100 5(x – 2) – 100 Solution: = 5x – 10 5x x – 100 – 90 – 16 = = = = Using the Distributive Law Add 10 to both sides Divide both sides by 5 Check: Using x= – 16 in the algebraic expression 5(x – 2) = 5(x – 2) – 100 This check proves we have a correct solution as we got the expected value of 2. Page | 6

  7. Olympic College Topic 2 – Solving Equations Example 4: Solve the equation 1.5(3x – 2) = 6 1.5(3x – 2) = Solution: 6 4.5x – 3 4. 5x x = = = = 6 9 2 Using the Distributive Law Add 3 to both sides Divide both sides by 4.5 Check: Using x= 2 in the algebraic expression 1.5(3x – 2) = 1.5(3(2) – 2) = 1.5(4) = 6 This check proves we have a correct solution as we got the expected value of 6. Example 5: Solve the equation (5x + 15) = 15 Solution: Method 1: (5x + 15) = 15 + = 15 Using the Distributive Law = Subtract from both sides = Divide both sides by x = 3 Solution: Method 2: (5x + 15) = 15 (5x + 15) = 5x + 15 = 2( 15) 30 Multiply both sides by 2 5x = = = 15 3 Subtract 15 from both sides Divide both sides by 5 x Check: Using x= 3 in the algebraic expression (5x + 15) = (5(3) + 15) = (30) = 15 This check proves we have a correct solution as we got the expected value of 15. Page | 7

  8. Olympic College Topic 2 – Solving Equations Solve the following equations:– Exercise 4A: 1. (a) 4(x + 2) = 32 (b) 2(2x – 6) = 20 (d) 5(4x – 2) (c) 2(x + 1) = 12 = 80 (f) 2(2x – 6) (h) 2(2y – 5) – 20 14 (e) 8(2y + 5) (g) 3(x + 2) = = 16 – 33 = = (4y – 4) (i) 5(4x + 1) = 20 (j) = 8 (k) 0.3(2x – 6) = – 30 (l) 10(2x + 1) = 10 (n) – 3(2x – 5) = (m) (3x + 1) = 16 21 (o) 1.2(1 – 2x) = (p) 0.05(4x – 2) = 24 10 (q) – 2(2x + 5) = (r) 8(2d – 6) – 24 16 = Page | 8

  9. Olympic College Topic 2 – Solving Equations Type 5: Complex Mixed Equations. There are many different forms that linear equations take, in order to solve them we use the balance method to change the form of the equation into its standard form ax + b = c where a,b and c are real numbers. There are a number of strategies that can be used to solve complex equations, the most common are Step 1: Step 2: Step 3: Step 4: Step 5: If there is a parenthesis then use the distributive law to remove them. Simplify each side of the equation by rearranging them and combining like terms. Use the balance method to simplify the equation to the standard form ax + b = c Use the balance method to rearrange the equation to the form ax = b Divide both sides of the equation ax = b by a to get the solution x = . Example 1: Solve the equation 12x + 5 = 7x + 25 Solution: 12x + 5 = 7x + 25 12x = 5x = = x = 7x + 20 20 4 Subtract 8 from both sides Subtract 7x from both sides Divide both sides by 5 Check: Using x= 4 in the algebraic expression 12x + 5 12(4) + 5 48 + 5 53 = = = = 7x + 25 7(4) + 25 28 + 25 53 This check proves we have a correct solution as both sides equal 53. Page | 9

  10. Olympic College Topic 2 – Solving Equations Example 2: Solve the equation 14x + 2 = 2x + 32 – 4x 2x + 34 – 4x Solution: 14x + 2 = 2x – 4x + 34 – 2x + 34 7x + 32 32 2 14x + 2 14x + 2 16x 16x x = = = = = = Rearrange like terms together Subtract 2 from both sides Subtract 7x from both sides Divide both sides by 16 2x + 34 – 4x Check: Using x= 2 we get 14x + 2 = 2(2) + 34 – 4(2) 4 + 34 – 8 30 14(2) + 2 28 + 2 30 = = = This check proves we have a correct solution as both sides equal 30. Example 3: Solve the equation Solve the equation 4x – 2 = 14 – 2x 4x – 2 14 – 2x Solution: = 16 – 2x 16 4x 6x = = = = Add 2 to both sides Add 2x to both sides Divide both sides by 6 x 4x – 2 – 2 – 2 14 – 2x 14 – 2( 14 – – Check: Use x = = = = = 4( – = Since both sides have a value of the solution is correct. Page | 10

  11. Olympic College Topic 2 – Solving Equations Example 4: Solve the equation 3x – 2 + 6x = 2x – 7 + 10x 3x – 2 + 6x 2x – 7 + 10x Solution: = 9x + 6x– 2 9x– 2 9x – 3x x 2x + 10x – 7 12x – 7 12x – 5 – 5 = = = = = = Rearrange like terms together Using the Distributive Law Add 2 to both sides Subtract 12x from both sides Divide both sides by – 3 x – 2 + 6x = 2x – 7 + 10x Check: Using x = into 3( ) – 2 + 6( = 2( ) – 7 + 10( 5 – 2 + 10 – = + 13 = 13 = 13 This check proves we have a correct solution as both sides equal 13. Example 5: Solve the equation 7.9x – 3.2 = 1.2(x – 3) + 3.5x 7.9x – 3.2 7.9x – 3.2 7.9x – 3.2 7.9x – 3.2 7.9x 3.2x 1.2(x – 3) + 3.5x 1.2x – 3.6 + 3.5x 3.5x + 1.2x – 3.6 4.7x – 3.6 4.7x – 0.4 – 0.4 Solution: = = = = = = = – 0.125 x = Check: Using – 1.125 7.9x – 3.2 1.2(x – 3) + 3.5x into = 7.9(– 0.125) – 3.2 – 0.9875 – 3.2 – 4.1875 1.2(– 0.125 – 3) + 3.5(– 0.125) 1.2(– 3.125) – 0.4375 – 4.1875 = = = This check proves we have a correct solution as both sides equal – 4.1875 Page | 11

  12. Olympic College Topic 2 – Solving Equations = x – 4 Example 6: Solve the equation x + x – 4 x – 4 – x – Solution: Method 1 x+ = x = x = Subtract from both sides x – x = x = – – Subtract from both sides 6(– – 26 6( x) x = = ) Multiply both sides by 6 x – 4 Solution: Method 2 x+ = 6( x – 4) 6( x + )= = Multiply both sides by 6 (LCD 2,3) = = Subtract Subtract from both sides from both sides x – 4 Check: Using x = into x+ = – 4 ( + = ( – 13 + = = This check proves we have a correct solution as both sides equal Example 7: Solve the equation 4(2x + 5) + 8 = 36 . Solution: 4(2x + 5) + 8 = 36 8x + 20 + 8 8x + 28 = = 8x = = x = 36 36 8 1 Using the Distributive Law Subtract 28 from both sides Divide both sides by 8 Check: Using x= 1 in 4(2x + 5) + 8 = 4(2(1) + 5) + 8 = 4(2 + 5) + 8 = 4(7) + 8 = 36 This check proves we have a correct solution as we get the expected result of 36. Page | 12

  13. Olympic College Topic 2 – Solving Equations Example 8: Solve the equation 6(2x + 2) = 2(x – 7) 2(x – 7) Solution: 6(2x + 2) = 2x – 14 2x – 26 – 26 12x + 12 12x 10x = = = = = Using the Distributive Law Subtract 12 from both sides Subtract 2x from both sides Divide both sides by 10 x = 2(x – 7) Check: Using x= into 6(2x + 2) – 7) – 7) 6(2( 6( + 2) + 2) = = 2( 2( – ) 6( + ) ) = = 2( 2( ) 6( = This check proves we have a correct solution as both sides equal Example 9: Solve the equation 4(3x + 1) + x = 3(x – 1) – 3x + 33 . 3(x – 1) – 3x + 33 Solution: 4(3x + 1) + x = 3x – 3 – 3x + 33 3x – 3x – 3 + 33 30 26 12x + 4 + x 12x + x + 4 13x + 4 13x = = = = = = Using the Distributive Law Collect like terms Subtract 4 from both sides Divide both sides by 13 x 3(x – 1) – 3x + 33 3(2 – 1) – 3(2) + 33 3(1) – 6 + 33 3 – 6 + 33 Check: Using x = into 4(3x + 1) + x 4(3(2) + 1) + 2 4(6 + 1) + 2 4(7) + 2 = = = = 30 = 30 This check proves we have a correct solution as both sides equal Page | 13

  14. Olympic College Topic 2 – Solving Equations Example 10: Solve the equation 5(x – 2) + 6 = 25 – 10(3x – 7) + 2x 5(x – 2) + 6 25 – 10(3x – 7) + 2x Solution: = 5x – 10 + 6 25 – 30x + 70 + 2x = Using the Distributive Law 5x – 10 + 6 5x – 4 – 30x + 2x + 70 + 25 Rearrange like terms – 28x + 95 = = – 28x + 99 99 3 5x 33x x = = = = Add 4 to both sides Add 28x to both sides Divide both sides by 33 5(x – 2) + 6 5(3– 2) + 6 5(1) + 6 11 11 11 25 – 10(3x – 7) + 2x 25 – 10(3(3) – 7) + 2(3) 25 – 10(9 – 7) + 6 25 – 10(2) + 6 25 – 20 + 6 11 Check: Using x = 3 in = = = = = = This check proves we have a correct solution as we got the expected value of 11. Page | 14

  15. Olympic College Topic 2 – Solving Equations Exercise 5A: Solve the following equations:– 1. (a). 5x + 3 = 2x + 9 (c). 4x + 9 = 2x + 13 (e). 15c + 4 = 4c + 15 (g). 5x – 3 + x = 3x + 9 – 12 (i). 8x = 3 – 2x + 17 (k). 5r – 12 – r = r (b). 9x + 5 = 4x + 15 (d). 8a + 3 = 5a + 21 (f). 3s + 3 = 8s + 33 (h). 9x – 5 = 4x + 15 + x (j). 13c – 9 = 6c + 33 + 13c (l). – 7 = – 3 + 2p + 10 Solve the following equations:– 2. (b). x – = x + 2 (d). 8f – = 3f + (f). 3.2 – 0.2v + 2 = v + 0.4 (h). 0.04x – 0.07 = 0.02x + 0.17 (a). x + 5 = x + 25 (c). y + 2 = y + 20 (e). 1.5s – 1.3 = 0.5s + 2 (g). 0.8n + 2.4 = 1.6n + 1.4 Solve the following equations if possible:– 3. (a) 1 + 3(x + 2) = (c) 2(2x + 5) + 6 = 2x + 2 4x + 8 (b) 2 + 2(2x + 3) = (d) 3(2x – 6) – 9 = 10 33 (e) 14x + 7 = 3x – 4 (g) 4(2x + 5) + 8 = 36 (f) 2x – 5(x+3) = 3x – 6 (h) 6(2x + 2) = 2(x – 7) 10(2y – 1) (i) 3(x + 2) = 2(x + 2) (j) 12(2y + 3) = (k) 2(2 – 2x) + 6 = (l) 2(2 – 4x) – 9 = 33 – x 4x + 8 (m) 14x + 7 = 7(21 – 3x) – 14 (0) 4(2b + 5) + 1 = 7b + 21 (n) 2 – 4(2x+2) = 6x – 16 (p) 6(2x + 2) = 2(6x – 7) Page | 15

  16. Olympic College Topic 2 – Solving Equations Solve the following equations if possible:– (a) 2(2x + 1) + 4x = 4(x – 1) – 5x + 28 4. (b) 3(5 – x) – 6 = 5 – 2(2x + 3) + 6x (c) – 4(4x + 1) + 8 = 2 + 7(2x – 1) – 21 (d) 4(4x – 2) + 8 = 23 – (3x – 7) – 30 (e) 4(4x – 2) + 8 = 23 – (3x – 7) – 30 + 11x (f) 3 + 2(x + 1) + 3 = 5 – (3x – 7) – 6 (g) 3(x + 1) + 2(3x + 1) = 3(x – 1) – 3(x + 3) – 1 (h) – (x – 2) + 6x = – 6(6 – x) + 41 What is the solution to the equation 5x – 4 = 12 – 3x 5. A. x = 1 B. x = 2 C. x = 4 D. x = 8 What is the solution to the equation 4(3x – 8) – 11x = 2 6. A. x = 6 B. x = 10 C. x = 30 D. x = 34 What is the solution to the equation 6(4.5b – 12) = 9 7. A. b = 28 B. b = 3 C. b = D. b = What is the solution to the equation 3p – 1 = 5(p – 1) – 2(7 – 2p) 8. C. p = – 9 D. p = – 1 A. p = 3 B. p = 0 Page | 16

  17. Olympic College Topic 2 – Solving Equations Solutions: Exercise 1A: 1. x = 7 6. k =13 11. x = 71 16. h = – 12 21. k = – 4 2. x = 10 7. t = 14 12. r = 87 17. u = – 26 22. k = – 1 3. x = 7 8. y = 3 13. = – 43 18. m = 11 23. k = 16 4. x = 2 9. e = 3 14. d = – 13 19. s = 32 24. x = 0 5. d = 11 10. p = 10 15. y = – 13 20. b = 60 Exercise 2A: (e) x = – 3 (j) x = 2 1. (a) x = 3 (f) x = – 5 (b) x = 4 (g) b = – 3 (c) x = 12 (h) c = – 3 (d) x = 2 (i) d = – 5 = – 30 = –50 (k) x = 3 (p) x = 20 (l) x = 0 (q) = 50 (m) x = 10 (r) = 15 (n) (o) Solve the following equations:– Exercise 3A: 1. x = 7 7. p = 4 13. x = 2 19. c = – 9 25. d = 2. x = 7 8. x = 5 14. x = 5 20. b = 8 26. b = 10 3. x = 3 9. x = 2 15. c = 9 21. f = 9 27. x = 3 4. x = 5 10. r = 4 16. x = 5 22. x = 1 28. b = 0 5. y = 2 11. t = 3 17. d = 23. e = 29. b = 6. a = 4 12. y = 2 18. b = 9 24. x = 6 30. x = – 11 31. x = 2 Exercise 4A: 32. x = 21 1.(a) x = 6 1.(f) x = – 2 1.(e) x = – 47 1.(p) x = 1.(b) x = 8 1.(g) x = – 13 1.(f) x = 1.(q) x = 1.(c) x = 5 1.(h) y = 6 1.(m) x = 1.(r) d = 1.(d) x = 1.(i) x = 1.(n) x = – 1 1.(e) x = 1.(j) y = 7 1.(o) x = Page | 17

  18. Olympic College Topic 2 – Solving Equations Exercise 5A: 1.(f) s = – 6 1.(a) x = 2 1.(b) x = 2 1.(c) x = 2 1.(d) a = 3 1.(e) c = 1 1.(j) c = – 7 1.(l )p = – 7 1.(g) x = 0 1.(h) x = 4 1.(i) x = 2 1.(k) r = 4 2.(b) x = – 2.(f) v = 4 3.(b) x = 3.(f) x = – 3.(j) x = 3.(n) x = 2.(c) y = – 180 2.(g) n = 1.25 3.(c) No solution 3.(g) x = 1 3.(k) x = 3.(o) b = 2.(a) x = 80 2.(e) s = 1.5 3.(a) x = – 5 3.(e) x = 3.(i) x = 3.(m) x = 2.(d) f = 2.(h) x = 6 3.(d) x = 10 3.(h) x = 3.(l) x = 3.(p) No solution 4.(a) x = 3 4.(e) no solution 5.B 6.D 4.(b) x = 2 4.(f) x = – 1 8.B 4.(c) x = 1 4.(g) x = – 2 4.(d) 4.(h) x=0 x= – 3 7.B Page | 18

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