# How do you solve this system of equation: x + 3y = 1 and - 3x - 3y = - 15?

##### 2 Answers
Jul 27, 2018

The solutions are $x = 7$ and $y = - 2$

#### Explanation:

By substitution

$\left\{\begin{matrix}x + 3 y = 1 \\ - 3 x - 3 y = - 15\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 1 - 3 y \\ x + y = 5\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 1 - 3 y \\ 1 - 3 y + y = 5\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 1 - 3 y \\ 1 - 2 y = 5\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 1 - 3 y \\ 2 y = - 4\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 1 - 3 \cdot - 2 = 7 \\ y = - 2\end{matrix}\right.$

Or graphically

graph{(x+3y-1)(x+y-5)=0 [-5.704, 12.076, -4.195, 4.694]}

$x = 7 , \setminus \setminus \setminus y = - 2$

#### Explanation:

Given equations

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

$- 3 x - 3 y = - 15$

$x + y = 5 \setminus \ldots \ldots \ldots \left(2\right)$

Subtracting (2) from (1), we get

$x + 3 y - \left(x + y\right) = 1 - 5$

$2 y = - 4$

$y = - 2$

setting the value of $y$ in (1), we get

$x + 3 \left(- 2\right) = 1$

$x - 6 = 1$

$x = 7$

Hence the solution is $x = 7 , \setminus \setminus \setminus y = - 2$