# How do you solve the following system?:  y = 3/4x + 5 , x + 2y = 3

$x = - \frac{14}{5}$ and $y = \frac{29}{10}$

#### Explanation:

From the given equations
$y = \frac{3 x}{4} + 5$ and $x + 2 y = 3$

Using the second equation $x + 2 y = 3$
Let $x = 3 - 2 y$ then substitute in the first equation

first equation becomes

$y = \frac{3 \left(3 - 2 y\right)}{4} + 5$

multiply both sides by of the equation by 4 now to get rid of any fraction

$4 y = 4 \cdot \frac{3 \left(3 - 2 y\right)}{4} + 4 \cdot 5$

$4 y = \cancel{4} \cdot \frac{3 \left(3 - 2 y\right)}{\cancel{4}} + 4 \cdot 5$

$4 y = 9 - 6 y + 20$

$4 y + 6 y = 9 + 20$

$10 y = 29$
divide both sides by 10

$\frac{10 y}{10} = \frac{29}{10}$

$\frac{\cancel{10} y}{\cancel{10}} = \frac{29}{10}$

$y = \frac{29}{10}$

Solve now for the value of x using the second equation

$x = 3 - 2 y$ and $y = \frac{29}{10}$

$x = 3 - 2 \cdot \left(\frac{29}{10}\right)$
$x = 3 - \frac{29}{5}$
$x = \frac{15 - 29}{5}$

$x = - \frac{14}{5}$

God bless...I hope the explanation is useful..