Y = x^2+5x-3
y-x=2
What is the solution of the linear-quadratic system of equations
1 answer:
Answer:
(-5,-3)
(1,3)
Step-by-step explanation:
Solving with substitution
Set both equations to y = ...
y = x^2 + 5x - 3
----------------
y - x =2
y = x + 2
----------------
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)
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