How do you solve the following system using substitution?: 5x+2y=-4, -3x+y=20

May 7, 2018

$\left(x , y\right) \to \left(- 4 , 8\right)$

Explanation:

$5 x + 2 y = - 4 \to \left(1\right)$

$- 3 x + y = 20 \to \left(2\right)$

$\text{rearrange equation "(2)" to express y in terms of x}$

$\text{add "3x" to both sides}$

$\Rightarrow y = 20 + 3 x \to \left(3\right)$

$\textcolor{b l u e}{\text{substitute "y=20+3x" into equation }} \left(1\right)$

$5 x + 2 \left(20 + 3 x\right) = - 4 \leftarrow \text{distribute}$

$\Rightarrow 5 x + 40 + 6 x = - 4$

$\Rightarrow 11 x + 40 = - 4$

$\text{subtract 40 from both sides}$

$\Rightarrow 11 x = - 44$

$\text{divide both sides by 11}$

$\Rightarrow x = \frac{- 44}{11} = - 4$

$\text{substitute "x=-4" into equation } \left(3\right)$

$\Rightarrow y = 20 - 12 = 8$

$\text{the point of intersection } = \left(- 4 , 8\right)$

May 7, 2018

The solutions are {x=-4 ; y=8}

Explanation:

The equations are

$\left\{\begin{matrix}5 x + 2 y = - 4 \\ - 3 x + y = 20\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}5 x + 2 y = - 4 \\ y = 3 x + 20\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}5 x + 2 \left(3 x + 20\right) = - 4 \\ y = 3 x + 20\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}11 x + 40 = - 4 \\ y = 3 x + 20\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}11 x = - 44 \\ y = 3 x + 20\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = - 4 \\ y = 3 \times - 4 + 20 = 8\end{matrix}\right.$