Question

solve 48/x + y-6/x-y=10 and
15/x+y+4/x-y=9​

Answer 1

Answer:

(1) : ( x,y) = (12,3)

(2) : ( x,y) = ( 8,3)

Answer 2

$Given \: pair \: equations :$

$\frac{48}{x+y} - \frac{6}{x-y} = 10 \: --(1)$

$and \: \frac{15}{x+y} + \frac{4}{x-y} = 9 \: --(2)$

$If \: we \: substitute \: \frac{1}{x+y} = a \: and$

$\frac{1}{x-y} = b, we \:get \:the \: following$

$pair \:of: linear \:equations$

$48a - 6b = 10$

/* Dividing each term by 2 , we get */

$\implies 24a - 3b = 5 \: --(3)$

$and \: 15a + 4b = 9 \: ---(4)$

$Equation (3) \times 4 : 96a-12b = 20 \:--(5)$

$Equation (4) \times 3 : 45a+12b = 27\:--(6)$

/* Add Equations (1) and (2) , we get */

$141a = 47$

$\implies a = \frac{47}{141}$

$\implies a = \frac{1}{3} \: --(7)$

/* Substitute the value of 'a' in equation (4), */

$15 \times \frac{1}{3} + 4b = 9$

$\implies 5 + 4b = 9$

$\implies 4b = 9 - 5$

$\implies 4b = 4$

$\implies b = \frac{4}{4}$

$\implies b = 1$

$But , \frac{1}{x+y} =a = \frac{1}{3}$

$\implies x + y = 3 \: --(8)$

$But , \frac{1}{x-y} =b = 1\: --(9)$

$\implies x - y = 1 \: --(9)$

/* Add Equations (8) and (9) ,we get */

$2x = 4$

$\blue { x = 2 }$

/* Put x = 2 in equation (8), we get */

$\blue { y = 1 }$

Therefore.,

$\green { x = 2 \: and \: y = 1 }$

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