# How do you solve the following linear system  3 x-2y=4 , x+4y=3?

Feb 12, 2017

$y = \frac{5}{14}$
$x = \frac{11}{7}$

#### Explanation:

1) Rearrange your equations to make them easier to use. The easiest way to do this is to isolate ONE variable that has the coefficient of 1.

$3 x - 2 y = 4$

$x = - 4 y + 3$

2) Since you now know what the x variable is in terms of y, you can go ahead and substitute it into the first equation.

$3 \left(- 4 y + 3\right) - 2 y = 4$

3) Simplify

$3 \left(- 4 y + 3\right) - 2 y = 4$

$- 12 y + 9 - 2 y = 4$

$- 14 y + 9 = 4$

$- 14 y = - 5$

$y = \frac{5}{14}$

4) Substitute the solution you just got to get the second variable.

$x + 4 y = 3$

$x + 4 \left(\frac{5}{14}\right) = 3$

$x + \frac{10}{7} = 3$

$x = \frac{11}{7}$

Feb 12, 2017

$x = \frac{11}{7} = 1 \frac{4}{7}$ and $y = \frac{5}{14}$

#### Explanation:

$3 x - 2 y = 4$
$x + 4 y = 3$

Using the second equation, derive a value for $x$.

$x + 4 y = 3$

Subtract $4 y$ from each side.

$x = 3 - 4 y$

In the first equation, substitute $x$ with $\textcolor{red}{\left(3 - 4 y\right)}$.

$3 x - 2 y = 4$

$3 \textcolor{red}{\left(3 - 4 y\right)} - 2 y = 4$

Open the brackets and simplify. The product of a positive and a negative is a negative.

$9 - 12 y - 2 y = 4$

$9 - 14 y = 4$

Subtract $9$ from each side.

$- 14 y = - 5$

Divide both sides by $- 14$.

$y = \frac{5}{14}$

In the second equation, substitute $y$ with $\textcolor{b l u e}{\frac{5}{14}}$.

$x + 4 y = 3$

$x + 4 \left(\textcolor{b l u e}{\frac{5}{14}}\right) = 3$

Open the brackets and simplify.

$x + 2 \cancel{4} \times \frac{5}{7 \cancel{14}} = 3$

$x + \frac{10}{7} = 3$

Subtract $\frac{10}{7}$ from each side.

$x = 3 - \frac{10}{7}$

$x = \frac{21}{7} - \frac{10}{7}$

$x = \frac{11}{7} = 1 \frac{4}{7}$