# How do you solve the following system: -2x + 5y = 20, 2x-y=7 ?

May 23, 2017

#### Answer:

$\left(\frac{55}{8} , \frac{27}{4}\right)$

#### Explanation:

$\textcolor{red}{- 2 x} + 5 y = 20 \to \left(1\right)$

$\textcolor{red}{2 x} - y = 7 \to \left(2\right)$

$\text{ the system is probably best solved using "color(blue)"elimination method}$

$\text{since the x terms have the same numeric value but with}$
$\text{opposing signs, adding them will eliminate the x term}$

$\left(1\right) + \left(2\right) \text{ term by term}$

$\left(- 2 x + 2 x\right) + \left(5 y - y\right) = \left(20 + 7\right)$

$\Rightarrow 4 y = 27$

$\text{divide both sides by 4}$

$\frac{\cancel{4} y}{\cancel{4}} = \frac{27}{4}$

$\Rightarrow y = \frac{27}{4}$

$\text{substitute this value into either " (1)" or } \left(2\right)$

$2 x - \frac{27}{4} = 7 \leftarrow \text{ substituting in } \left(2\right)$

$\Rightarrow 2 x = 7 + \frac{27}{4} = \frac{55}{4}$

$\Rightarrow x = \frac{55}{8}$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values in ( 1 )

$- 2 \left(\frac{55}{8}\right) + 5 \left(\frac{27}{4}\right) = - \frac{55}{4} + \frac{135}{4} = 20 \rightarrow \text{ True}$

$\Rightarrow \text{point of intersection} = \left(\frac{55}{8} , \frac{27}{4}\right)$