# How do you find the solution of the system of equations y=-x^2+2x-3 and y=x-5?

May 2, 2015

We can substitute $y = x - 5$ in the first equation

We get $x - 5 = - {x}^{2} + 2 x - 3$

Transposing all the terms on the right to the left, we get

$\to x - 5 + {x}^{2} - 2 x + 3 = 0$

$\to {x}^{2} - x - 2 = 0$

To solve for x, we need to factorise the expression on the left

We can split the middle term of this expression to factorise it

$\to {x}^{2} - 2 x + x - 2 = 0$

$\to x \left(x - 2\right) + 1 \left(x - 2\right) = 0$

$\left(x - 2\right)$ is a common factor to both the terms

$\to \left(x - 2\right) \left(x + 1\right) = 0$

This tells us that:

Either $\left(x - 2\right) = 0$ or $\left(x + 1\right) = 0$

color(green)( x = 2 or x = -1 is the Solution