More linear and quadratic systems help?

y=x^2−3x+7

y=3x−2

Set yhe y's equal and we have that

x^2 - 3x + 7  =   3x - 2          subtract 3x / add 2   from/to both sides

x^2 - 6x + 9   =  0        factor

(x - 3)^2 = 0        which implies that x = 3

So.......plug 3 into either function  and we get that   y = 3(3) - 2   =  7    

Set it so that \(x^2-3x+7=3x-2\) .

Now bring over to the left side: \(x^2-6x+9=0\)

We can find out that \(x=3\).

Now plug it in to \(y=3x-2 => y=3(3)-2 => y=7\)

Therefore, the y-coordinate is 7.

You are very welcome!

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