How do you solve 6x+y=13 and y-x= -8 using substitution?

Aug 8, 2017

See a solution process below:

Explanation:

Step 1) Solve the second equation for $y$:

$y - x = - 8$

$y - x + \textcolor{red}{x} = - 8 + \textcolor{red}{x}$

$y - 0 = - 8 + x$

$y = - 8 + x$

Step 2) Substitute $\left(- 8 + x\right)$ for $y$ in the first equation and solve for $x$:

$6 x + y = 13$ becomes:

$6 x + \left(- 8 + x\right) = 13$

$6 x - 8 + x = 13$

$6 x - 8 + 1 x = 13$

$6 x + 1 x - 8 = 13$

$\left(6 + 1\right) x - 8 = 13$

$7 x - 8 = 13$

$7 x - 8 + \textcolor{red}{8} = 13 + \textcolor{red}{8}$

$7 x - 0 = 21$

$7 x = 21$

$\frac{7 x}{\textcolor{red}{7}} = \frac{21}{\textcolor{red}{7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} x}{\cancel{\textcolor{red}{7}}} = 3$

$x = 3$

Step 3) Substitute $3$ for $x$ in the solution to the second equation at the end of Step 1 and calculate $y$:

$y = - 8 + x$ becomes:

$y = - 8 + 3$

$y = - 5$

The Solution Is: $x = 3$ and $y = - 5$ or $\left(3 , - 5\right)$

Aug 8, 2017

$\left(x , y\right) \to \left(3 , - 5\right)$

Explanation:

$6 x + \textcolor{red}{y} = 13 \to \left(1\right)$

$\textcolor{red}{y} - x = - 8 \to \left(2\right)$

$\text{from } \left(2\right) \textcolor{w h i t e}{x} \textcolor{red}{y} = x - 8 \to \left(3\right)$

$\text{substitute "y=x-8" in } \left(1\right)$

$\Rightarrow 6 x + x - 8 = 13$

$\Rightarrow 7 x - 8 = 13$

$\text{add 8 to both sides}$

$7 x \cancel{- 8} \cancel{+ 8} = 13 + 8$

$\Rightarrow 7 x = 21$

$\text{divide both sides by 7}$

$\frac{\cancel{7} x}{\cancel{7}} = \frac{21}{7}$

$\Rightarrow x = 3$

$\text{substitute this value in "(3)" and evaluate y}$

$\Rightarrow y = 3 - 8 = - 5$

$\textcolor{b l u e}{\text{As a check}}$

$\text{substitute these values into } \left(1\right)$

$\left(6 \times 3\right) + \left(- 5\right) = 13 \leftarrow \text{ True}$

$\Rightarrow \text{point of intersection } = \left(3 , - 5\right)$
graph{(y+6x-13)(y-x+8)((x-3)^2+(y+5)^2-0.04)=0 [-16.02, 16.02, -8.01, 8.01]}