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

Jul 18, 2018

$x = - \frac{5}{3}$ and $y = - \frac{7}{3}$

#### Explanation:

Dividing by $2$, $2 x - 4 y = 6$ gives us $x - 2 y = 3$

or $x - y - y = 3$

Multiplying by $3$, we get

$3 x - 3 y - 3 y = 9$, but $3 x - 3 y = 2$,

therefore $2 - 3 y = 9$

or $2 - 9 = 3 y$ i.e. $3 y = - 7$ and $y = - \frac{7}{3}$

Hence $x - 2 y = 3$ becomes $x - \left(- \frac{14}{3}\right) = 3$

or $x = 3 - \frac{14}{3} = - \frac{5}{3}$

Hence $x = - \frac{5}{3}$ and $y = - \frac{7}{3}$

Jul 18, 2018

The solution is $S = \left\{\begin{matrix}x = - \frac{5}{3} \\ y = - \frac{7}{3}\end{matrix}\right.$

#### Explanation:

Solve the simutaneous equations by substitution

$\left\{\begin{matrix}2 x - 4 y = 6 \\ 3 x - 3 y = 2\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}2 x = 4 y + 6 \\ 3 x - 3 y = 2\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 2 y + 3 \\ 3 \left(2 y + 3\right) - 3 y = 2\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 2 y + 3 \\ 6 y + 9 - 3 y = 2\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 2 y + 3 \\ 3 y = - 9 + 2\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 2 \cdot \left(- \frac{7}{3}\right) + 3 \\ y = - \frac{7}{3}\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = - \frac{5}{3} \\ y = - \frac{7}{3}\end{matrix}\right.$