I will assume your second equation is
y = 2x + 1
sub that into the first:
2x + 1 = x^2 + 5x - 9
x^2 + 3x - 10 = 0
(x+5)(x-2) = 0
x = -5 or x = 2
if x = -5, y = 2(-5) + 1 = -9
if x = 2 , y = 5
so they intersect at (-5,-9) and (2,5)
What are the solutions to the system?
y=x^2+5x-9
y+2x+1
2 answers
Eq1: Y = x^2 + 5x - 9
Eq2: Y = 2x + 1 ?
In Eq1, replace Y with 2x + 1
2x + 1 = x^2 + 5x - 9
x^2 + 5x - 9 - 2x - 1 = 0
x^2 + 3x -10 = 0
-10 = -2*5. Sum = -2 + 5 = 3 = B.
(x-2)(x+5) = 0
x-2 = 0
X = 2
Y = 5.(Substitute 2 for x in Eq1 or Eq2)
x+5 = 0
X = -5
Y = -9(Substituted -5 for x in Eq1 or Eq2).
Solution Sets: (x, y) = (2, 5), (-5,-9).
Eq2: Y = 2x + 1 ?
In Eq1, replace Y with 2x + 1
2x + 1 = x^2 + 5x - 9
x^2 + 5x - 9 - 2x - 1 = 0
x^2 + 3x -10 = 0
-10 = -2*5. Sum = -2 + 5 = 3 = B.
(x-2)(x+5) = 0
x-2 = 0
X = 2
Y = 5.(Substitute 2 for x in Eq1 or Eq2)
x+5 = 0
X = -5
Y = -9(Substituted -5 for x in Eq1 or Eq2).
Solution Sets: (x, y) = (2, 5), (-5,-9).