# How do you solve the following system of equations?: 13x+13y=16, 3x + 19y=19?

Oct 22, 2017

$x = \frac{57}{208} , y = \frac{199}{208}$

#### Explanation:

$13 x + 13 y = 16$.....$\left(I\right)$

$3 x + 19 y = 19$.....$\left(I I\right)$

Rearranging $\left(I I\right)$ for x for simplicity,

$3 x = 19 - 19 y$

$\implies$$x = \frac{19 - 19 y}{3}$

Substituting for $x$ in $\left(I\right)$

$13 \left(\frac{19 - 19 y}{3}\right) + 13 y = 16$

$\implies$$\frac{247 - 247 y}{3} + 13 y = 16$

$\implies$$\frac{247 - 247 y + 39 y}{3} = 16$

$\implies$$247 - 208 y = 16 \cdot 3$

$\implies$$208 y = 247 - 48$

$\implies$$y = \frac{199}{208}$

Substituting for $y$ in either equation (we'll choose $\left(I I\right)$),

$3 x + 19 \left(\frac{199}{208}\right) = 19$

$\implies$$3 x = 19 - \frac{3781}{208}$

$\implies$$x = \frac{1}{3} \cdot \left(\frac{3952 - 3781}{208}\right) = \frac{1}{3} \cdot \left(\frac{171}{208}\right) = \frac{171}{624} = \frac{57}{208}$