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1 of 5 Skillsheet Solving equations by balancing Imagine an equation as a pan balance. x 1 4 5 8 x14 8 If we take 2 from the right-hand pan, it will no longer balance. x 1 4 ≠ 6 x14 6 If we add 2 to the right-hand pan it will not balance. x 1 4 ≠ 10 x14 10 It will only balance if we add or subtract the same amount from each pan. For example, subtract 2. x 1 2 5 6 x12 6 We can find out the value of x by taking all of the numbers from the left-hand pan and the same numbers from the right-hand pan to balance. x 5 4 x 4 © Cengage Learning Australia Pty Ltd 2013 MAT12AMSS00001 Algebra and Modelling: Further algebraic skills and techniques MAT12AMSS00001.indd 1 www.nelsonnet.com.au 07/05/13 3:42 PM 2 of 5 Solving equations Example 1 x2356 x59 Add 3 to both sides: x23 6 x 9 2x 3 Example 2 2x 5 3 05x13 23 5 x x 523 Add x to both sides: x13 Take 3 from both sides: 23 x 2x 6 x 3 x 2 4 x 8 Example 3 2x 5 6 x53 Halve both sides (or divide both sides by 2): Example 4 x 54 2 x58 Double both sides (multiply both sides by 2): Exercises 1 Solve each equation. a x 1 3 5 7 b e x 1 5 5 0 f 3x 5 12 i 2x 5 210 x 1 5 5 12 x j 5 8 3 c x 2 6 5 13 g 8x 5 48 k x 2 5 5 2 d x 2 3 5 23 h 25x 5 30 x l 2 5 6 2 © Cengage Learning Australia Pty Ltd 2013 MAT12AMSS00001 Algebra and Modelling: Further algebraic skills and techniques MAT12AMSS00001.indd 2 www.nelsonnet.com.au 07/05/13 3:42 PM 3 of 5 Solving equations requiring more steps Often we need to continue to balance the equation to find the value of x. Example 5 2x 1 7 5 17 2x17 17 2x 5 10 Take 7 from both sides: 2x 10 x55 Divide both sides by 2: x 5 2(x 1 3) 10 Example 6 2(x 1 3) 5 10 x1355 Divide both sides by 2: x13 5 x52 Take 3 from both sides: x 2 Example 7 3x 1 2 5 x 1 10 3x 1 2 x 1 10 Take x from both sides: 2x 1 2 10 2x 5 8 Take 2 from both sides: 2x 8 x54 Divide both sides by 2: x 4 2x 1 2 5 10 Exercises 2 Solve each equation. a 6x 2 7 5 17 b 5 1 2x 5 17 c 5x 2 3 5 213 e 7 2 2x 5 1 f 4 2 3x 5 22 i j 4x 1 5 5 2x 1 13 x g 1 4 5 8 4 k 5x 1 2 5 x 1 10 n 5x 2 2 5 7x 2 12 o 6x 1 13 5 4x 1 13 x 2 3 5 27 2 m x 2 3 5 4x 2 9 d 6 2 3x 5 9 x 2 4 5 27 5 l x 1 6 5 6x 2 9 h © Cengage Learning Australia Pty Ltd 2013 MAT12AMSS00001 Algebra and Modelling: Further algebraic skills and techniques MAT12AMSS00001.indd 3 www.nelsonnet.com.au 07/05/13 3:42 PM 4 of 5 Regrouping and balancing Sometimes it is easier to change the grouping of the terms before balancing the equation. Example 8 11 2(x 2 4)1(x 1 4) 2(x 2 4) 1 (x 1 4) 5 11 2x 2 8 1 x 1 4 5 11 3x 2 4 11 Add 4 to both sides: 3x 15 Divide both sides by 3: x 5 3x 2 4 5 11 3x 5 15 x55 Exercises 3 Solve each equation. a 2(2x 1 3) 1 2(x 1 4) 5 20 c 3(3x 2 4) 2 2(2x 2 3) 5 211 e 2(x 2 7) 5 6(x 1 1) b 3(x 1 3) 1 2(x 2 5) 5 4 d 3(x 1 4) 5 2(4x 1 1) f 4(x 1 3) 5 22(x 1 6) Example 9 2x 2 6 5 22 5 2x 26 5 22 2x 5 4 Multiply both sides by 5: 2x 20 Divide both sides by 2: x 10 Add 6 to both sides: 2x 54 5 2x 5 20 x 5 10 Exercises 4 'Solve each equation.' 2x 2 6 5 2 5 x12 e 5 2 4 3x 1 1 i 5 10 4 2 2 5x m 5 23 6 a 3x 1 3 5 15 4 x 15 f 5 1 2 2 x 14 j 5 4 5 7 2 4x n 53 9 b 4x 1 5 5 1 9 x24 g 523 8 8x 2 5 k 523 7 c 4x 2 11 5 9 3 x 27 522 h 5 3x 2 4 l 52 4 d © Cengage Learning Australia Pty Ltd 2013 MAT12AMSS00001 Algebra and Modelling: Further algebraic skills and techniques MAT12AMSS00001.indd 4 www.nelsonnet.com.au 07/05/13 3:42 PM 5 of 5 Answers 1 a x 5 4 e x 5 25 i x 5 10 2 a x 5 4 e x 5 3 i x 5 28 m x 5 2 3 a x 5 1 e x 5 25 4 a x 5 20 e x 5 6 i x 5 13 m x 5 4 b x 5 7 f x 5 4 j x 5 24 b x 5 6 f x 5 2 j x 5 4 n x 5 5 b x 5 1 f x 5 24 b x 5 16 f x 5 23 j x 5 8 n x 5 25 c x 5 19 g x 5 6 k x 5 210 c x 5 22 g x 5 16 k x 5 2 o x 5 0 c x 5 21 d x 5 0 h x 5 26 l x 5 212 d x 5 21 h x 5 215 l x 5 3 c x 5 29 g x 5 220 k x 5 22 d x 5 15 h x 5 23 l x 5 4 d x52 © Cengage Learning Australia Pty Ltd 2013 MAT12AMSS00001 Algebra and Modelling: Further algebraic skills and techniques MAT12AMSS00001.indd 5 www.nelsonnet.com.au 07/05/13 3:42 PM