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Simultaneous equations 1. The method of elimination 2. The method of substitution 1 of 43 © Boardworks Ltd 2005 The elimination method – Example 1 If two equations are true for the same values, we can add or subtract them to give a third equation that is also true for the same values. For example, suppose 3x + y = 9 5x – y = 7 Adding these equations: 3x + y = 9 + 5x – y = 7 8x divide both sides by 8: 2 of 43 The y terms have been eliminated. = 16 x=2 © Boardworks Ltd 2005 The elimination method – Example 1 continued Adding the two equations eliminated the y terms and gave us a single equation in x. Solving this equation gave us the solution x = 2. To find the value of y when x = 2 substitute this value into one of the equations. 3x + y = 9 5x – y = 7 Substituting x = 2 into the first equation gives us: 3×2+y=9 6+y=9 subtract 6 from both sides: 3 of 43 y=3 © Boardworks Ltd 2005 The elimination method – Example 1 continued We can check whether x = 2 and y = 3 solves both: 3x + y = 9 5x – y = 7 by substituting them into the second equation. 5×2–3=7 10 – 3 = 7 This is true, so we have confirmed that x=2 y=3 solves both equations. 4 of 43 © Boardworks Ltd 2005 The elimination method – Example 2 Solve these equations: 3x + 7y = 22 3x + 4y = 10 Subtracting gives: 3x + 7y = 22 – 3x + 4y = 10 3y = 12 y=4 divide both sides by 3: The x terms have been eliminated. Substituting y = 4 into the first equation gives us, 3x + 7 × 4 = 22 3x + 28 = 22 3x = –6 subtract 28 from both sides: x = –2 divide both sides by 3: 5 of 43 © Boardworks Ltd 2005 The elimination method – Example 2 continued We can check whether x = –2 and y = 4 solves both, 3x + 7y = 22 3x + 4y = 10 by substituting them into the second equation. 3 × –2 + 7 × 4 = 22 –6 + 28 = 22 This is true and so, x = –2 y=4 solves both equations. 6 of 43 © Boardworks Ltd 2005 The elimination method – Practice 1 7 of 43 © Boardworks Ltd 2005 The elimination method – Example 3 Sometimes we need to multiply one or both of the equations before we can eliminate one of the variables. For example, 4x – y = 29 1 3x + 2y = 19 2 We need to have the same number in front of either the x or the y before adding or subtracting the equations. Call these equations 1 and 2 . 2× 1 : 3 + 2 : divide both sides by 11: 8 of 43 8x – 2y = 58 + 3x + 2y = 19 11x 3 = 77 x=7 © Boardworks Ltd 2005 The elimination method - Example 3 continued To find the value of y when x = 7 substitute this value into one of the equations, 1 4x – y = 29 3x + 2y = 19 2 Substituting x = 7 into 1 gives, 4 × 7 – y = 29 28 – y = 29 subtract 28 from both sides: multiply both sides by –1: –y = 1 y = –1 Check by substituting x = 7 and y = –1 into 2 , 3 × 7 + 2 × –1 = 19 21 – 2 = 19 9 of 43 © Boardworks Ltd 2005 The elimination method - Example 4 Solve: 2x – 5y = 25 1 3x + 4y = 3 2 Call these equations 1 and 2 . 3× 1 2× 2 3 – 4 , 6x – 15y = 75 – 6x + 8y = 6 divide both sides by –23: 3 4 – 23y = 69 y = –3 Substitute y = –3 in 1 , 2x – 5 × –3 = 25 2x + 15 = 25 2x = 10 subtract 15 from both sides: x=5 divide both sides by 2: 10 of 43 © Boardworks Ltd 2005 The elimination method – Practice 2 11 of 43 © Boardworks Ltd 2005 The substitution method – Example 1 Two simultaneous equations can also be solved by substituting one equation into the other. For example, y = 2x – 3 2x + 3y = 23 1 2 Call these equations 1 and 2 . Substitute equation 1 into equation 2 . 2x + 3(2x – 3) = 23 2x + 6x – 9 = 23 expand the brackets: 8x – 9 = 23 simplify: 8x = 32 add 9 to both sides: x=4 divide both sides by 8: 12 of 43 © Boardworks Ltd 2005 The substitution method – Example 1 continued To find the value of y when x = 4 substitute this value into one of the equations, 1 y = 2x – 3 2x + 3y = 23 2 Substituting x = 4 into 1 gives y=2×4–3 y=5 Check by substituting x = 4 and y = 5 into 2 , 2 × 4 + 3 × 5 = 23 8 + 15 = 23 This is true and so the solutions are correct. 13 of 43 © Boardworks Ltd 2005 The substitution method – Example 2 How could the following pair of simultaneous equations be solved using substitution? 3x – y = 9 8x + 5y = 1 1 2 One of the equations needs to be arranged in the form x = … or y = … before it can be substituted into the other equation. Call these equations 1 and 2 . Rearrange equation 1 . add y to both sides: subtract 9 from both sides: 14 of 43 3x – y = 9 3x = 9 + y 3x – 9 = y y = 3x – 9 © Boardworks Ltd 2005 The substitution method – Example 2 continued 3x – y = 9 8x + 5y = 1 1 2 Now substitute y = 3x – 9 into equation 2 . 8x + 5(3x – 9) = 1 8x + 15x – 45 = 1 expand the brackets: 23x – 45 = 1 simplify: 23x = 46 add 45 to both sides: x=2 divide both sides by 23: Substitute x = 2 into equation 1 to find the value of y. 3×2–y=9 6–y=9 –y = 3 y = –3 15 of 43 © Boardworks Ltd 2005 The substitution method – Example 2 continued 3x – y = 9 1 8x + 5y = 1 2 Check the solutions x = 2 and y = –3 by substituting them into equation 2 . 8 × 2 + 5 × –3 = 1 16 – 15 = 1 This is true and so the solutions are correct. Solve these equations using the elimination method to see if you get the same solutions for x and y. 16 of 43 © Boardworks Ltd 2005