

InterviewSolution
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Pick out the solution from the values given in the bracket next to each equation. show that the other values do not satisfy the equation.(a) 5m = 60 (10, 5, 12, 15)(b) n + 12 = 20 (12, 8, 20, 0)(c) p – 5 = 5 (0, 10, 5 – 5)(d) q/2 = 7 (7, 2, 10, 14)(e) r – 4 = 0 (4, -4, 8, 0)(f) x + 4 = 2 (-2, 9, 2, 4) |
Answer» (a) 5m = 60 (10, 5, 12, 15) m = 12 a solution to the given equation because form = 12 5m = 5 × 12 = 60 and hence, the equation is satisfied m = 10 is not a solution to the given equation because for m = 10. 5m = 5 × 10 = 50, and net 60 m = 5 is net a solution to the given equation because for m = 5, 5m = 5 × 5 = 25, and net 60 m = 15 is net a solution to the given equation because for m = 15 5m = 5 × 15 = 75, and net 6. (b) n + 12 = 20 (12, 8, 20, 0) n = 8 is a solution to the given equation because for n = 8 n + 12 = 8 + 12 = 20 and hence, the equation is satisfied n +12 = 12 + 12 = 24, and net 20 n = 20 is net a solution to the given equation because for n = 20 n + 12 = 20 + 12 = 32, and net 20 (c) p – 5 = 5 (0, 10, 5 – 5) p = 10 is a solution to the given equation because for p = 10, p – 5 = 10 – 5 = 5 and hence, the equation is satisfied. p = 0 is net a solution to the given equation because for p = 0 . p – 5 = 5 – 5 = 0, and net 5 p = -5 is net s solution to the given equation because for p = -5 (d) q/2 = 7 (7, 2, 10, 14) q = 14 is a solution to the given equation because for q = 7 (q/2) = 7/2 and not 7 q/2 = 7/2 and net 7 q = 2 is not a solution to the given equation because for q = 7 q/2 = 7/2 and net 7 q = 2 is net a solution to the given equation because for q = 2 q/2 = 2/2 = 1, and net 7 q = 10 is net a solution to the given equation because for qn = 10, q/2 = 10/2 = 5, and net 7 (e) r – 4 = 0 (4, -4, 8, 0) r = 4 is a solution to the given equation because for r = 4 r = 4 = 4 – 4 = 0 and hence, the equation is satisfied r = -4 is net a solution to-the given equation because for r = -4 r – 4 = -4 – 4 = -8 and net 0 r = 8 is net a solution to the given equation because for r = 8 r – 4 = 8 – 4 = 4 and net 0 r = 0 is net a solution to the given equation because for r = 0 r – 4 = 0 – 4 = -4 and net 0 . (f) x + 4 = 2 (-2, 9, 2, 4) x = -2 is a solution to the given equation because for x – 2 x + 4 = -2 + 4 = 2 and hence, the equation is satisfied x = 0 is net a solution to the given equation because for x = 0 x + 4 = 0 + 4 = 4 and net 2 x = 2 is net a solution to the given equation because for x = 2 x + 4 = 2 + 4 = 6 and net 2 x = 4 is net a solution to the given equation because for x = 4 x + 4 = 4 + 4 = 8, and net 2. |
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