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

Dec 19, 2017

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

Explanation:

$19 \cdot \left(13 x + 13 y\right) - 13 \cdot \left(3 x + 19 y\right) = 19 \cdot 16 - 13 \cdot 19$

$247 x + 247 y - 39 x - 247 y = 304 - 247$

$208 x = 57$, so $x = \frac{57}{208}$

Hence,

$3 \cdot \frac{57}{208} + 19 y = 19$

$\frac{171}{208} + 19 y = 19$

$19 y = 19 - \frac{171}{208}$

$19 y = \frac{3781}{208}$, so $y = \frac{199}{208}$