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

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Answer 1 · 2015-05-02T07:28:59

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

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

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

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

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

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

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

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

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

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

This tells us that:

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

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

How do you find the solution of the system of equations y=−x2+2x−3y=-x^2+2x-3 and y=x−5y=x-5?