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Solve the following systems of equations:\(\frac{44}{x+y}\)+\(\frac{30}{x-y}\) = 10\(\frac{55}{x-y}\)+\(\frac{40}{x-y}\) = 13 |
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Answer» \(\frac{44}{x+y}\)+\(\frac{30}{x-y}\) = 10 \(\frac{55}{x-y}\)+\(\frac{40}{x-y}\) = 13 Multiplying eq1 by 5 and eq2 by 4 and subtracting ⇒ - 10/(x – y) = - 2 ⇒ x – y = 5 Multiply eq1 by 4 and eq2 by 3 and subtracting ⇒ 11/(x + y) = 1 ⇒ (x + y) = 11 Thus, 2x = 16 ⇒ x = 8 ∴ y = 8 – 5 = 3 |
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