SOLUTION: Hello tutor, I really need your help to solve this question please. Solve the system: y = x^2 + 7, y = 4x + 3 Thank you

Question 971554: Hello tutor, I really need your help to solve this question please.

Solve the system: y = x^2 + 7, y = 4x + 3

Thank you
Found 2 solutions by checkley77, Edwin McCravy:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
y = x^2 + 7
y = 4x + 3
x^2+7=4x+3
x^2-4x+7-3=0
x^2+4x+4=0
(x+2)(x+2)=0
x-2=0
x=-2 ans.
y=-2^2+7
y=4+7=11 ans
y=4*2+3=8+3=11 ans.
Answer by Edwin McCravy(20055) (Show Source): You can put this solution on YOUR website!
y = x^2 + 7, y = 4x + 3

Since y equals to both x^2 + 7 and to 4x + 3, set them equal to each other:
x^2 + 7 = 4x + 3
Get 0 on the right by subtracting the right side from both sides:
x^2 - 4x + 4 = 0
Factor the left side:
(x - 2)(x - 2) = 0
x - 2 = 0; x - 2 = 0
x = 2 x = 2
There is one solution since both came out the same.
Find y by substituting in either of the two original equations:
y = 4x + 3
y = 4(2) + 3
y = 8 + 3
y = 11
So the solution is (x,y) = (2,11), which means that the graphs of
y = x^2 + 7 and y = 4x + 3
intersect at the point (2,11)

Edwin
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