# How do you solve the system of equations y-3 = (x-1)^2 and y-x=4 algebraically?

Apr 27, 2017

The Soln. Set $= \left\{\begin{matrix}0 & 4 \\ 3 & 7\end{matrix}\right\} .$

#### Explanation:

$y - x = 4 \Rightarrow y = x + 4.$

Subst.ing, in, $y - 3 = {\left(x - 1\right)}^{2} ,$ we have,

$x + 4 - 3 = {\left(x - 1\right)}^{2} , \text{ i.e., } x + 1 = {x}^{2} - 2 x + 1 ,$ or,

${x}^{2} - 3 x = 0.$

$\therefore x \left(x - 3\right) = 0 \Rightarrow x = 0 , \mathmr{and} , x = 3.$

 y=x+4, &, x=0 rArr y=4, and,

 y=x+4, &, x=3 rArr y=7.

These roots satisfy the given eqns.

Hence, the Soln. Set $= \left\{\begin{matrix}0 & 4 \\ 3 & 7\end{matrix}\right\} .$

Apr 27, 2017

$x = 0$ or $x = 3$ and $y = 4$ or $y = 7$

#### Explanation:

You can rearrange the second equation to get just $y$
$y = x + 4$
the you can rearrange the first equation to get all the unknown numbers on one side
$0 = {\left(x - 1\right)}^{2} - y + 3$
then expand
$0 = {x}^{2} - 2 x + 4 - y$

the second equation is equal to $y$ therefore we can substitute this equation into the first equation everywhere where there is a $y$

$0 = {x}^{2} - 2 x + 4 - \left(x + 4\right)$
then solve
$0 = {x}^{2} - 2 x + 4 - x - 4$
$0 = {x}^{2} - 3 x$
therefore $x = 0$ or $x = 3$
substitute back into original equations to find y

$y = x + 4$
$y = 4$ or $y = 7$

Apr 27, 2017

$\left(0 , 4\right) , \left(3 , 7\right)$

#### Explanation:

Simplify the first equation to $y = {\left(x - 1\right)}^{2} + 3 = {x}^{2} - 2 x + 4$
We can then plug this into the second equation to get
${x}^{2} - 2 x + 4 - x = 4$

Then simplify
${x}^{2} - 3 x + 4 = 4$
${x}^{2} - 3 x = 0$

Then use the quadratic formula to solve
$\setminus \frac{- \left(- 3\right) \setminus \pm \setminus \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(0\right)}}{2 \left(1\right)}$
$\setminus \frac{3 \setminus \pm \setminus \sqrt{9 - 0}}{2}$
$\setminus \frac{3 \setminus \pm 3}{2}$

Solving we get
$\setminus \frac{3 + 3}{2} = \frac{6}{2} = 3$
$\setminus \frac{3 - 3}{2} = \frac{0}{2} = 0$