# How do you solve the following linear system: y = 4x - 13 , 31x - 6y = 29 ?

Jul 19, 2017

See a solution process below:

#### Explanation:

Step 1) Because the first equation is already solved for $y$ we can substitute $\left(4 x - 13\right)$ for $y$ in the second equation and solve for $x$:

$31 x - 6 y = 29$ becomes:

$31 x - 6 \left(4 x - 13\right) = 29$

$31 x - \left(6 \cdot 4 x\right) + \left(6 \cdot 13\right) = 29$

$31 x - 24 x + 78 = 29$

$\left(31 - 24\right) x + 78 = 29$

$7 x + 78 = 29$

$7 x + 78 - \textcolor{red}{78} = 29 - \textcolor{red}{78}$

$7 x - 0 = - 49$

$7 x = - 49$

$\frac{7 x}{\textcolor{red}{7}} = - \frac{49}{\textcolor{red}{7}}$

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

$x = - 7$

Step 2) Substitute $- 7$ for $x$ in the first equation and calculate $y$:

$y = 4 x - 13$ becomes:

$y = \left(4 \times - 7\right) - 13$

$y = - 28 - 13$

$y = - 41$

The Solution Is: $x = - 7$ and $y = - 41$ or $\left(- 7 , - 41\right)$