# How do you solve the following system: 7y − 6x − 5 = 0, y + 4x = 16 ?

May 21, 2017

See a solution process below:

#### Explanation:

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

$y + 4 x = 16$

$y + 4 x - \textcolor{red}{4 x} = 16 - \textcolor{red}{4 x}$

$y + 0 = 16 - 4 x$

$y = 16 - 4 x$

Step 2) Substitute $\left(16 - 4 x\right)$ for $y$ in the first equation and solve for $x$:

$7 y - 6 x - 5 = 0$ becomes:

$7 \left(16 - 4 x\right) - 6 x - 5 = 0$

$\left(7 \cdot 16\right) - \left(7 \cdot 4 x\right) - 6 x - 5 = 0$

$112 - 28 x - 6 x - 5 = 0$

$112 - 5 - 28 x - 6 x = 0$

$\left(112 - 5\right) + \left(- 28 - 6\right) x = 0$

$107 + \left(- 34\right) x = 0$

$107 + \left(- 34\right) x + \textcolor{red}{34 x} = 0 + \textcolor{red}{34 x}$

$107 + 0 = 34 x$

$107 = 34 x$

$\frac{107}{\textcolor{red}{34}} = \frac{34 x}{\textcolor{red}{34}}$

$\frac{107}{34} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{34}}} x}{\cancel{\textcolor{red}{34}}}$

$\frac{107}{34} = x$

$x = \frac{107}{34}$

Step 3) Substitute $\frac{107}{34}$ for $x$ in the solution to the second equation at the end of Step 1 and calculate $y$:

$y = 16 - 4 x$ becomes:

$y = 16 - \left(4 \times \frac{107}{34}\right)$

$y = \left(\frac{34}{34} \times 16\right) - \frac{428}{34}$

$y = \frac{544}{34} - \frac{428}{34}$

$y = \frac{116}{34}$

$y = \frac{2 \times 58}{2 \times 17}$

$y = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 58}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 17}$

$y = \frac{58}{17}$

The solution is: $x = \frac{107}{34}$ and $y = \frac{58}{17}$ or $\left(\frac{107}{34} , \frac{58}{17}\right)$