InterviewSolution
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Solve the following equations without transposing and check your result. (i) x + 5 = 9 (ii) y – 12 = -5 (iii) 3x + 4 = 19 (iv) 9z = 81 (v) 3x + 8 = 5x + 2 (vi) 5y + 10 = 4y – 10 |
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Answer» i) x + 5 = 9 x + 5 – 5 9 – 5 (subtract 5 from both sides) x = 4 Check LHS = x + 5 (substituting x = 4) = 4 + 5 = 9 RHS = 9 ∴ L.H.S = R.H.S ii) y – 12 = – 5 y – 12 = – 5 y – 12 + 12= – 5 + 12 (add l2on both sides) y = 7 Check LHS = y – 12 = 7 – 12= – 5 RHS = -5 ∴ L.H.S = R.H.S iii) 3x+4= 19 3x + 4 = 19 3x + 4 – 4 = 19 – 4 (subtract 4 from both sides) 3x = 15 \(\frac {3x}{3} = \frac {15}{3}\) (Divide both sides by 3) x = 5 Check LHS = 3x + 4 = 3 x 5 + 4 = 15 + 4 = 19 RHS = 19 ∴ L.,H.S = R.H.S iv) 9z = 81 \(\frac {9z}{9} = \frac {81}{9}\) (Divide both sides by 9) z = 9 Check LHS = 9z = 9 x 9 = 81 RHS = 81 ∴ LHS = RHS v) 3x + 8 = 5x + 2 3x + 8 = 5x + 2 3x + 8 – 8 = 5x + 2 – 8 (Adding -8 on both sides) 3x = 5x – 6 3x – 5x = 5x – 6 – 5x (Subtract 5x from both sides) -2x = -6 \(\frac {-2x}{-2} = \frac {-6}{-2}\) (Divide both sides by -2) Check LHS = 3x + 8 = 3(3) + 8 = 9 + 8 = 17 RHS = 5x + 2 = 5(3) + 2 = 15 + 2 = 17 ∴ LHS = RHS (vi) 5y + 10 = 4y – 10 5y + 10 = 4y – 10 5y + 10 – 1o = 4y – 10 – 10 (Subtract 10 from both sides) 5y =4y – 20 5y – 4y = 4y – 20 – 4y (Substract ty from both sides) y = – 20 Check LHS = 5y + 10 = 5 x( – 20) + 10 = – 100+ 10= – 90 RHS = 4y – 10 = 4 x ( – 20) – 10 = – 80 – 10 = -90 ∴ LHS = RHS |
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