# What is the solution of the linear-quadratic system of equations? y=x^2+5x-3, y?x=2

## Question:

What is the solution of the linear-quadratic system of equations?

{eq}y = x^2 + 5x - 3 \\ y - x = 2 {/eq}

## Consistent System of Equations:

In order for a system of equations to be consistent they should have a solution that is common for the both of them. This is a set of coordinates that will make both equations true. Graphically, it would be the coordinates of the point or points they intersect.

The system of equations being considered are:

{eq}y = x^2 + 5x - 3 \ \ \rm (Eq. 1) \\ \it y - x = 2 \ \ \rm (Eq. 2) {/eq}

One solution would be by substitution. From Eq. 2, an identity for {eq}y {/eq} would be:

{eq}\begin{align} y - x &= 2 \\ y = x + 2 \ \ \rm (Eq. 3) \end{align} {/eq}

Substituting Eq. 3 in Eq. 1 and solving for the possibel values of {eq}x {/eq}:

{eq}\begin{align} y &= x^2 + 5x - 3 \\ x + 2 &= x^2 + 5x - 3 \\ 0 &= x^2 + 5x - 3 -x -2 \\ 0 &= x^2 +4x -5 \end{align} {/eq}

Factoring:

{eq}(x +5 )(x - 1)= 0 {/eq}

In order for this equation to be true then either factor would have to be zero. To find the values of {eq}x {/eq} that would make this true is:

{eq}\begin{align} x + 5 &= 0 \\ x &= -5 \\ \\ x -1 &= 0 \\ x &= 1 \end{align} {/eq}

Using these values in Eq. 3 to find the values of {eq}y {/eq}:

{eq}\begin{align} x = -5 \\ y = x + 2 \\ y = -5 + 2 \\ y = -3 \\ \\ x = 1 \\ y = x + 2 \\ y = 1 + 2 \\ y = 3 \end{align} {/eq}

The solutions for the linear-quadratic system of equations {eq}y = x^2 + 5x - 3 \ \ \rm and \it y - x = 2 \ \ \rm {/eq} are the points (-5, -3) and (1,3). 