# How do solve the following linear system?:  -3x -8y = -9 , 4 x+ 16y = 2 ?

Sep 3, 2017

$\textcolor{red}{\left(x , y\right) = \left(8 , - \frac{15}{8}\right)}$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} - 3 x - 8 y = - 9$
[2]$\textcolor{w h i t e}{\text{XXX}} 4 x + 16 y = 2$

Multiplying [1] by $2$
[3]$\textcolor{w h i t e}{\text{XXX}} - 6 x - 16 y = - 18$

Adding [2] and [3]
$\textcolor{w h i t e}{\text{XXX+(-")4x+16y=color(white)("xxxx}} 2$
$\textcolor{w h i t e}{\text{XXX}} + \underline{- 6 x - 16 y = - 18}$
[4]$\textcolor{w h i t e}{\text{XXX")-2xcolor(white)("xxxxx}} = - 16$

Dividing both sides of [4] by $\left(- 2\right)$
[5]$\textcolor{w h i t e}{\text{XXX}} x = 8$

Substituting $8$ for $x$ in [1]
[6]$\textcolor{w h i t e}{\text{XXX}} \left(- 3\right) \times 8 - 8 y = - 9$
$\textcolor{w h i t e}{\left[6\right]}$$\textcolor{w h i t e}{\text{XXX}} - 24 - 8 y = - 9$
$\textcolor{w h i t e}{\left[6\right]}$$\textcolor{w h i t e}{\text{XXX}} - 8 y = 15$
[7]$\textcolor{w h i t e}{\text{XXX}} y = - \frac{15}{8}$

So we have $\left(x , y\right) = \left(8 , - \frac{15}{8}\right)$

We can verify this result by substituting these values for $x$ and $y$ in equation [2]

Sep 3, 2017

See a solution process below:

#### Explanation:

**Step 1) Solve both of the equations for $8 y$:

Equation 1:

$- 3 x - 8 y = - 9$

$\textcolor{red}{3 x} - 3 x - 8 y = \textcolor{red}{3 x} - 9$

$0 - 8 y = 3 x - 9$

$- 8 y = 3 x - 9$

$\textcolor{red}{- 1} \times - 8 y = \textcolor{red}{- 1} \left(3 x - 9\right)$

$8 y = \left(\textcolor{red}{- 1} \times 3 x\right) - \left(\textcolor{red}{- 1} \times 9\right)$

$8 y = - 3 x + 9$

Equation 2:

$4 x + 16 y = 2$

$\frac{4 x + 16 y}{\textcolor{red}{2}} = \frac{2}{\textcolor{red}{2}}$

$\frac{4 x}{\textcolor{red}{2}} + \frac{16 y}{\textcolor{red}{2}} = 1$

$2 x + 8 y = 1$

$- \textcolor{red}{2 x} + 2 x + 8 y = - \textcolor{red}{2 x} + 1$

$0 + 8 y = - 2 x + 1$

$8 y = - 2 x + 1$

Step 2) Because the left side of each equation are equal we can now equate the right side of each equation and solve for $x$:

$- 3 x + 9 = - 2 x + 1$

$\textcolor{red}{3 x} - 3 x + 9 - \textcolor{red}{1} = \textcolor{red}{3 x} - 2 x + 1 - \textcolor{red}{1}$

$0 + 8 = \left(\textcolor{red}{3} - 2\right) x + 0$

$8 = 1 x$

$8 = x$

$x = 8$

Step 3) Substitute $8$ for $x$ into the solution for either equation in Step 1 and solve for $y$:

$8 y = - 2 x + 1$ becomes:

$8 y = \left(- 2 \times 8\right) + 1$

$8 y = - 16 + 1$

$8 y = - 15$

$\frac{8 y}{\textcolor{red}{8}} = - \frac{15}{\textcolor{red}{8}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} y}{\cancel{\textcolor{red}{8}}} = - \frac{15}{8}$

$y = - \frac{15}{8}$

The Solution Is: $x = 8$ and $y = - \frac{15}{8}$ or $\left(8 , - \frac{15}{8}\right)$