Question

Solve for x and y .10/2x+y + 3/2x-y = 3 and 15/2x+y + 9/2x- y = 6​

Answer 1

Answer:

$\boxed{x = 2, \: y = 1}$

Step-by-step explanation:

To solve the linear equations

$\frac{10}{2x + y} + \frac{3}{2x - y} = 3 \\ \\ \frac{15}{2x + y} + \frac{9}{2x - y} = 6 \\ \\ let \\ \\ \frac{1}{2x + y} = a \\ \\ \frac{1}{2x - y} = b \\ \\ so \\ \\ 10a + 3b = 3...eq1 \\ \\ 15a + 9b = 6 \\ \\ or \\ \\ 5a + 3b = 2 \: \: \: ...eq2$

Solve eq1 and eq2 by elimination method

$10a + 3b = 3 \\ 5a + 3b = 2 \\ (- ) \: \: ( - ) \: \: ( - ) \\ - - - - - - - \\ 5a = 1 \\ \\ a = \frac{1}{5} \\ \\ put \: value \: of \: a \: in \: eq2 \\ \\ 5( \frac{1}{5} ) + 3b = 2 \\ \\ 3b = 2 - 1 \\ \\ 3b = 1 \\ \\ b = \frac{1}{3}$

Now put these values of a and b in our assumption

$\frac{1}{2x + y} = \frac{1}{5} \\ \\ \frac{1}{2x - y} = \frac{1}{3} \\ \\ on \: cross \: multiplying \\ \\ 2x + y = 5 \\ 2x - y = 3 \\ - - - - - - \\ 4x = 8 \\ \\ x = 2 \\ \\ 2(2) + y = 5 \\ \\ 4 + y = 5 \\ \\ y = 1$

Hope it helps you.