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+1 vote
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in Linear Equations by (50.7k points)
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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

1 Answer

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Best 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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