Question

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

Answer 1

Answer:

(-5,-3)

(1,3)

Step-by-step explanation:

Solving with substitution

Set both equations to y =  ...

y = x^2 + 5x - 3

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y - x =2

y = x + 2

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Substitute one of the Y's with the other Y = equation.

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

Set one side to 0

0 = x^2 + 4x -5

x^2 + 4x - 5 = 0

Simplify by factoring

(x-1) (x+5)

Get your X values

x-1 = 0

x = 1

x+5 = 0

x = -5

Plug in each X into one of the equations to get your Y value.

1st X

y = x + 2

y = 1 + 2

y = 3

When X = 1, Y = 3

2nd X

y = x + 2

y = -5 + 2

y = -3

When X = - 5, Y = -3

Your answers are (1,3) and (-5,-3)