# How do solve the following linear system?: 7x + 1 = -3y, 3x+ 2y - 3 = 0 ?

$x = - \frac{11}{5} , \setminus \setminus \setminus y = \frac{24}{5}$

#### Explanation:

Given equations

$7 x + 1 = - 3 y$

$7 x + 3 y = - 1 \setminus \ldots \ldots . . \left(1\right)$ &

$3 x + 2 y - 3 = 0$

$3 x + 2 y = 3 \setminus \ldots \ldots . . \left(2\right)$

Multiplying (2) by $\frac{3}{2}$ & subtracting from (1) as follows

$7 x + 3 y - \frac{3}{2} \left(3 x + 2 y\right) = - 1 - \frac{3}{2} \left(3\right)$

$\frac{5}{2} x = - \frac{11}{2}$

$x = - \frac{11}{5}$

setting value of $x$ in (1), we get

$7 \left(- \frac{11}{5}\right) + 3 y = - 1$

$3 y = \frac{72}{5}$

$y = \frac{24}{5}$