Question

10/x+y+2/x-y=4;15x+y-5x-y=-2​

Answer 1

Step-by-step explanation:

This problem of pair of linear equation in 2 variables can be solved by the above mentioned method....and the methods used are substitution and elimination.....

hope it helps.......

Answer 2

Step-by-step explanation:

Given:

$\frac{10}{x + y} + \frac{2}{x - y} = 4 \: \: \: ...eq1\\ \\ \frac{15}{x + y} - \frac{5}{x - y} = - 2 \: \: \: ...eq2$

To find : Solution of equation.

Solution:

Step 1: Convert given equations in linear equations in two variables.

let

$\frac{1}{x - y} = a \: \: \: ...eq3 \\ \\ \frac{1}{x + y} = b \: \: \: ...eq4$

$\bold{\red{10a + 2b = 4}} \: \: \: ...eq5 \\ \bold{\green{15a - 5b = - 2}} \: \: \: ...eq6$

Step 2: Solve eq5 and eq6 and find values of a and b.

Multiply eq5 by 5 and eq6 by 2 and add both equations

$50a + 10b = 20 \\ 30a - 10b = - 4 \\ - - - - - - - - \\ 80a = 16 \\ \\ a = \frac{16}{80} \\ \\ \bold{\purple{a = \frac{1}{5}}}$

put value of a in eq5

$10 \times \frac{1}{5} + 2b = 4 \\ \\ 2 + 2b = 4 \\ \\ 2b = 4 - 2 \\ \\ 2b = 2 \\ \\\bold{\pink{ b = 1}}$

Step 3: Put the value of a and b in eq3 and eq4

$\frac{1}{x + y} = \frac{1}{5} \\ \\ or \\ \\ \bold{\green{x + y = 5}} \: \: \: ...eq7 \\ \\ \frac{1}{x - y} = \frac{1}{1} \\ \\ or \\ \\\bold{\red{ x - y = 1}} \: \: \: ...eq8$

add eq7 and eq8

$x + y = 5 \\ x - y = 1 \\ - - - - - - - \\ 2x = 6 \\ \\ x = \frac{6}{2} \\ \\ \bold{\red{x = 3}}$

put value of x in eq7

$3 + y = 5 \\ \\ y = 5 - 3 \\ \\ \bold{\green{y = 2 }}$

Final answer:

Value of x and y are

$\bold{\red{x = 3}} \\ \\ \bold{\green{y = 2 }}$

Hope it helps you.

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