How do you solve the following system: 11y + 3x = 4, -6y - 2x = 0 -3 ?

Jul 25, 2016

The Soln. $x = \frac{9}{4} , y = - \frac{1}{4}$.

Explanation:

$11 y + 3 x = 4. \ldots \ldots \ldots \ldots \ldots \left(1\right)$

$- 6 y - 2 x = - 3. \ldots \ldots \ldots . \left(2\right)$.

Multiplying $\left(1\right)$ by $2$, we get, $22 y + 6 x = 8. . . \ldots \left(1 \star\right)$

Multiplying $\left(2\right)$ by $3$, we get, $- 18 y - 6 x = - 9. . . \ldots \left(2 \star\right)$

Adding $\left(1 \star\right) \mathmr{and} \left(2 \star\right)$, we get,

$22 y \cancel{+ 6 x} - 18 y \cancel{- 6 x} = 8 - 9$

$\therefore 4 y = - 1 \Rightarrow y = - \frac{1}{4}$.

Sub.ing $y = - \frac{1}{4}$ in $\left(2\right)$, we get,

$- 6 \left(- \frac{1}{4}\right) - 2 x = - 3 , i . e . , - 2 x = - 3 - \frac{3}{2} = - \frac{9}{2}$

$\therefore x = \frac{9}{4}$.

Hence, the Soln. $x = \frac{9}{4} , y = - \frac{1}{4}$.