# How do you solve the system of equations algebraically y=6-x, x=4.5+y?

Jan 19, 2017

$\left(\frac{21}{4} , \frac{3}{4}\right)$

#### Explanation:

We can substitute $y = 6 - x \text{ into } x = 4.5 + y$

$\Rightarrow x = 4.5 + \left(\textcolor{red}{6 - x}\right) \leftarrow \text{ one variable}$

$\Rightarrow x = 4.5 + 6 - x$

$\Rightarrow 2 x = 10.5$

dividing by 2

$\Rightarrow x = 5.25 = \frac{21}{4}$

Substitute this value into y = 6 - x

$x = \frac{21}{4} \to y = 6 - \frac{21}{4} = \frac{3}{4}$

$\Rightarrow \left(\frac{21}{4} , \frac{3}{4}\right) \text{ is the solution}$

This could also have been solved by substituting

$x = 4.5 + y \text{ into " y=6-x}$

You may wish to attempt this yourself, as a check.