Question 971554
 y = x^2 + 7, y = 4x + 3
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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)

{{{graph(400,400,-13,13,-5,21,x^2+7,4x+3)}}}

Edwin</pre>