# How do solve the following linear system?:  8x + 3y= -9 , x - 3y = 1 ?

May 2, 2017

See the entire solution process below:

#### Explanation:

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

$x - 3 y = 1$

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

$x - 0 = 1 + 3 y$

$x = 1 + 3 y$

Step 2) Substitute $1 + 3 y$ for $x$ in the first equation and solve for $y$:

$8 x + 3 y = - 9$ becomes:

$8 \left(1 + 3 y\right) + 3 y = - 9$

$\left(8 \cdot 1\right) + \left(8 \cdot 3 y\right) + 3 y = - 9$

$8 + 24 y + 3 y = - 9$

$8 + 27 y = - 9$

$- \textcolor{red}{8} + 8 + 27 y = - \textcolor{red}{8} - 9$

$0 + 27 y = - 17$

$27 y = - 17$

$\frac{27 y}{\textcolor{red}{27}} = - \frac{17}{\textcolor{red}{27}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{27}}} y}{\cancel{\textcolor{red}{27}}} = - \frac{17}{27}$

$y = - \frac{17}{27}$

Step 3) Substitute $- \frac{17}{27}$ for $y$ in the solution to the second equation at the end of Step 1 and calculate $x$:

$x = 1 + 3 y$ becomes:

$x = 1 + \left(3 \times - \frac{17}{27}\right)$

$x = 1 - \frac{17}{9}$

$x = \left(\frac{9}{9} \times 1\right) - \frac{17}{9}$

$x = \frac{9}{9} - \frac{17}{9}$

$x = - \frac{8}{9}$

The solution is: $x = - \frac{8}{9}$ and $y = - \frac{17}{27}$ or $\left(- \frac{8}{9} , - \frac{17}{27}\right)$

May 2, 2017

x = -8/9; y = -17/27

#### Explanation:

$8 x + 3 y = - 9 \text{ }$ Equation 1

$x - 3 y = 1 \text{ }$ Equation 2

Eliminate $y$ by adding the two equations. The result becomes

$9 x = - 8$

Hence,

$x = - \frac{8}{9}$

Find y by substituting the already known $x$ to any of the original equation (or both just to double-check)

Using Equation 1,

$8 \left(- \frac{8}{9}\right) + 3 y = - 9$

$- 64 + 27 y = - 81$

$27 y = - 17$

Hence,

$y = - \frac{17}{27}$

To double-check, use also Equation 2 to get $y$

$- \frac{8}{9} - 3 y = 1$

$- 8 - 27 y = 9$

$- 17 = 27 y$

Hence,

$y = - \frac{17}{27}$