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

To determine the solution of the given system, we can use substitution method. We do as follows:

y=x^2+5x-3

y-x=2

Substitute y.

x^2+5x-3-x=2

Add common factors.

x^2+4x-3=2

Subtract 2 from both sides of the equation.

x^2+4x-5=0

Quadratic Formula:

[-4±√4² - (4*1*-5) ] / (2*1) = 0

[-4±√16 - (-20) ]/2=0

[-4±√36]/2=0

[-4±6]/2=0

(-4+6) / 2

2/2

x=1

(-4-6) / 2

-10/2

x=-5

Answer:

x=1

or

x=-5