# How do you solve [(x^2+1, 5-y), (x+y,y-4)]=[(5,x), (5,3)]?

Jan 18, 2017

$\left(x , y\right) = \left(- 2 , 7\right)$

#### Explanation:

$\left[\left({x}^{2} + 1 , \textcolor{w h i t e}{\text{x"),5-y),(x+y,,y-4)]=[(5,color(white)("x}} , x\right) , \left(5 , , 3\right)\right]$

$\Rightarrow$
$\textcolor{w h i t e}{\text{XX}} {x}^{2} + 1 = 5$
$\textcolor{w h i t e}{\text{XX}} 5 - y = x$
$\textcolor{w h i t e}{\text{XX}} x + y = 5$
$\textcolor{w h i t e}{\text{XX}} y - 4 = 3$

Examining the above,
we note that both  and  only involve a single variable
and of the two  would appear to be the easiest to solve:
$y - 4 = 3 \textcolor{w h i t e}{\text{XX")rarrcolor(white)("XX}} y = 7$

We can now substitute $7$ for $y$ back into 
$5 - 7 = x \textcolor{w h i t e}{\text{XX")rarrcolor(white)("XX}} x = - 2$

We now have our solution $\left(x , y\right) = \left(- 2 , 7\right)$
provided this is consistent with the other two equations:
$\textcolor{w h i t e}{\text{XX}} {x}^{2} + 1$ with $x = - 2 \textcolor{w h i t e}{\text{x")rarrcolor(white)("x}} {2}^{2} + 1 = 5$ (correct)
$\textcolor{w h i t e}{\text{XX}} x + y = 5$ with $x = - 2$ and $y = 7 \textcolor{w h i t e}{\text{x")rarrcolor(white)("X}} - 2 + 7 = 5$ (correct)

So $\left(x , y\right) = \left(- 2 , 7\right)$ is a consistent (valid) solution.