# How do you solve -x-3y= 15 and 2x+7y= -36 using substitution?

May 22, 2017

$x = 3$

#### Explanation:

-x - 3y = 15" "("Solve for "x)

$3 y = 15 + - x$

$- x = 15 + 3 y$

$\frac{- x}{-} 1 = \frac{15 + 3 y}{-} 1$

$x = \frac{15 + 3 y}{-} 1$

$\frac{15 + 3 y}{-} 1 \times \frac{- 1}{-} 1 = \frac{- 15 + - 3 y}{1}$

$2 x + 7 y = - 36$

Substitute $- 15 + - 3 y$ for x

$2 \cdot \left(- 15 + - 3 y\right) + 7 y = - 36 \text{(Distribute)}$

$\left(2 \cdot - 15\right) + \left(2 \cdot 3 y\right) + 7 y = - 36$

$- 30 + - 6 y + 7 y = - 36$

$- 30 + y = - 36$

$y = - 6$

Plug in the y's value

$- x - 3 \cdot - 6 = 15$

$- x - \left(- 18\right) = 15$

$- x + 18 = 15$

$- x = 15 - 18$

$- x = - 3$

$- \left(- x\right) = - \left(- 3\right)$

$x = 3$

But an easier approach would be using elimination method.

May 22, 2017

3,-6)

#### Explanation:

$- \textcolor{red}{x} - 3 y = 15 \to \left(1\right)$

$2 \textcolor{red}{x} + 7 y = - 36 \to \left(2\right)$

$\text{from " (1)" we can make x the subject}$

$\Rightarrow \textcolor{red}{x} = - 3 y - 15 \to \left(3\right)$

$\text{substitute this value into } \left(2\right)$

$\Rightarrow 2 \left(- 3 y - 15\right) + 7 y = - 36$

$\Rightarrow - 6 y - 30 + 7 y = - 36 \leftarrow \text{ distributing}$

$\Rightarrow y - 30 = - 36 \leftarrow \text{ simplifying left side}$

$\text{add 30 to both sides}$

$y \cancel{- 30} \cancel{+ 30} = - 36 + 30$

$\Rightarrow y = - 6$

$\text{substitute into " (3)" and evaluate for x}$

$x = \left(- 3 \times - 6\right) - 15 = 18 - 15 = 3$

$\Rightarrow \text{point of intersection } = \left(3 , - 6\right)$