# How do you solve using elimination of 3x+6y=12 and 2x-2.5y=13?

Nov 7, 2015

The system has one dolution: $\left\{\begin{matrix}x = 5 \frac{7}{13} \\ y = - \frac{10}{13}\end{matrix}\right.$

#### Explanation:

The original system is:

$\left\{\begin{matrix}3 x + 6 y = 12 \\ 2 x - 2.5 y = 13\end{matrix}\right.$

The elimination method means to change the system so that it is possible to eliminate one variable by adding (or substracting) both sides of the equations:

First we can divide both sides of the first equation by $3$ to make all the coefficients smaller and multiply the second equation by $2$ to make all the coefficients integer:

$\left\{\begin{matrix}x + 2 y = 4 \\ 4 x - 5 y = 26\end{matrix}\right.$

Now if we multiply the first equation by $\left(- 4\right)$ the coefficients of $x$ will be opposite numbers ($4$ and $- 4$)

$\left\{\begin{matrix}- 4 x - 8 y = - 16 \\ 4 x - 5 y = 26\end{matrix}\right.$

Now if we add both sides of the equations the variable $x$ will be eliminated:

$- 13 y = 10$

$y = - \frac{10}{13}$

Now if we move $2 y$ to the right side of the first equation before the last multiplication we get:

$x = 4 - 2 y$

Now we only have to put calculated $y$ in this equation and calculate $x$

$x = 4 - 2 \cdot \left(- \frac{10}{13}\right)$

$x = 4 + \frac{20}{13}$

$y = \frac{52 + 20}{13}$

$y = \frac{72}{13}$

$y = 5 \frac{7}{13}$