Question

solve faster using elimination method

Answer 1

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Given that :
$\frac{44}{x + y} + \frac{30}{x - y} = 10 \\ \\ \frac{55}{x + y} + \frac{40}{x - y} = 13$

$putting \: \: \\ \frac{1}{x + y} = a \: \: and \: \: \frac{1}{x - y} = b \: \: we \: \: get$

44a + 30b = 10 .....(i)

55a + 40b = 13 ......(ii)

Now, multiplying (i) and (ii) by 55 and 44 respectively, we get :

2420a + 1650b = 550

2420a + 1760b = 572

On subtraction, we get :

110b = 22

=> b = 1/5

From (ii), we get :

55a + 40×(1/5) = 13

=> 55a + 8 = 13

=> 55a = 5

=> a = 1/11

We assumed :

$\frac{1}{x + y} = a = \frac{1}{11} \\ \\ and \\ \\ or \: \: x + y = 11 \\ \\ and \\ \\ \frac{1}{x - y} = b = \frac{1}{5} \\ \\ or \: \: x - y = 5$

Thus, we have :

x + y = 11 .....(iii)

x - y = 5 .....(iv)

On addition, we get :

2x = 16

=> x = 8

From (iii), putting x = 8, we get :

8 + y = 11

=> y = 3

Therefore, the required solution is

x = 8 and y = 3.

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