# How do you solve x+y=3 and x = 4-(y-1)^2?

Apr 9, 2016

$\left(x , y\right) = \left(0 , 3\right)$
or
$\left(x , y\right) = \left(3 , 0\right)$

#### Explanation:

If $x + y = 3$
then $y = 3 - x$

and we can substitute $\left(\textcolor{red}{3 - x}\right)$ in place of $y$ in the second equation: $x = 4 - {\left(y - 1\right)}^{2}$

This gives
$\textcolor{w h i t e}{\text{XXX}} x = 4 - {\left(\textcolor{red}{3 - x} - 1\right)}^{2}$

$\textcolor{w h i t e}{\text{XXX}} x = 4 - {\left(2 - x\right)}^{2}$

$\textcolor{w h i t e}{\text{XXX}} x = 4 - \left(4 - 4 x + {x}^{2}\right)$

#color(white)("XXX")x= 4x-x^2

$\textcolor{w h i t e}{\text{XXX}} 3 x - {x}^{2} = 0$

$\textcolor{w h i t e}{\text{XXX}} x \left(3 - x\right) = 0$

$\textcolor{w h i t e}{\text{XXX")rarr x=0color(white)("XX")orcolor(white)("XX}} x = 3$

If $x = 0$ then $x + y = 3 \rightarrow y = 3$

If $x = 3$ then $x + y = 3 \rightarrow y = 0$