3x + 5y = 10 and 2x - 5y = 10 has solution

Given:

3x + 5y = 10 ....(1)

2x - 5y = 10  .....(2)

Adding eq (1) and eq (2), we get:

$\Rightarrow$3x + 5y + 2x - 5y = 10 + 10

$\Rightarrow$5x = 20

$\Rightarrow$x = 4

Substitute value of x = 4 in eq (1), we get:

$\Rightarrow$3x + 5y = 10

$\Rightarrow$3(4) + 5y = 10

$\Rightarrow$12 + 5y = 10

$\Rightarrow$ 5y = 10 - 12

$\Rightarrow$ 5y = - 2

$\Rightarrow$ y = $\frac{-2}{5}$

Therefore, $x=4 , y=\frac{-2}{5}$ are solution of 3x + 5y = 10 and 2x - 5y = 10.