Question

5/x+y+2/x-y=2and 4/x+y-3/x-y=14 find the solution ​

Answer 1

Question :-

Given above ↑

Answer :-

$\rightarrow \: \: \boxed{ \sf{ x = \frac{ 161}{1054} }} \: \: \: \: \: \boxed{ \sf{y \: = \frac{552}{1054} }} \:$

Solution :-

Given equations are ↓

$\rightarrow \sf{ \frac{5}{x + y} + \frac{2}{x - y} = 2} \: \: \: .......eq.(1) \\ \\ \rightarrow \sf{ \frac{4}{x + y} - \frac{3}{x - y} = 14} \: \: \: \: ...eq.(2) \\ \\ \implies \: 4 \times \sf{ eq. (1) - 5 \times \: eq. (2)} \\ \\ \rightarrow \sf{ \frac{8}{x - y} + \frac{15}{x - y} = - 62} \\ \\ \rightarrow \sf{ \frac{23}{x - y} = - 62} \\ \\ \rightarrow \boxed{\sf{x - y = - \frac{23}{62} \: \: \: ......eq.(3)}} \\ \\ \sf{now \: \: \: 3 \times \: eq.(1) + 2 \times \: eq.(2)} \\ \\ \rightarrow \sf{ \frac{15}{x + y} + \frac{8}{x + y} = 6 + 28} \\ \\ \rightarrow \sf{ \frac{23}{x + y} = 34} \\ \\ \rightarrow \boxed{ \sf{ x + y = \frac{23}{34} \: \: \: ....eq.(4)}} \\ \\ \sf{now \: adding \: eq.(3) \: \: and \: \: eq.(4)} \\ \\ \rightarrow \sf{ 2x = \frac{23}{34} - \frac{23}{62} } \\ \\ \rightarrow \sf{ 2x = \frac{731 - 391}{1054} } \\ \\ \rightarrow \boxed{ \sf{ x = \frac{ 161}{1054} }} \\ \sf{put \: this \: x \: in \: eq.(3)} \\ \\ \rightarrow \sf{ \frac{161}{1054} - y = \frac{ - 23}{62} } \\ \\ \rightarrow \sf{ y \: = \frac{23}{62} + \frac{161}{1054} } \\ \\ \rightarrow \sf{y \: = \frac{391 +161}{1054} } \\ \\ \rightarrow \boxed{ \sf{y \: = \frac{552}{1054} }}$

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