Question 1197265
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The variable y is equal to both x^2-7x+15 and also to 2x-5
Equate the right hand sides and get everything to one side.


x^2-7x+15 = 2x-5
x^2-7x+15-2x+5 = 0
x^2-9x+20 = 0


Then we can factor. Think of two numbers that
A) multiply to 20, and
B) add to -9
Through trial and error you should arrive at -4 and -5


-4 times -5 = 20
-4 plus -5 = -9


Therefore,
x^2-9x+20 = 0
(x-4)(x-5) = 0
x-4 = 0 or x-5 = 0
x = 4 or x = 5
The second to last step used the zero product property


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Alternative Route


Go back to x^2-9x+20 = 0
Compare this to ax^2+bx+c = 0


We have
a = 1
b = -9
c = 20
Plug those values into the quadratic formula
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-9)+-sqrt((-9)^2-4(1)(20)))/(2(1))}}}


{{{x = (9+-sqrt(81 - 80))/(2)}}}


{{{x = (9+-sqrt(1))/(2)}}}


{{{x = (9+-  1)/(2)}}}


{{{x = (9+1)/(2)}}} or {{{x = (9-1)/(2)}}}


{{{x = (10)/(2)}}} or  {{{x = (8)/(2)}}}


{{{x = 5}}} or  {{{x = 4}}}
The order of the solutions doesn't matter. 


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Once we've determined the x values of the solutions, we use them to find their paired y values.


Plug in x = 4
Let's do so for the first equation
y = x^2-7x+15
y = 4^2-7*4+15
y = 16-28+15
y = -12+15
y = 3
Be sure to follow the order of operations PEMDAS


This basically says the input x = 4 leads to the output y = 3


Now do the same for the other equation
y = 2x-5
y = 2*4-5
y = 8-5
y = 3
The second equation is much easier to work with, so if you only had to pick one, then I'd go for this.


However, it's good practice to check BOTH equations to verify the solution fully.


One solution is (x,y) = (4,3) which is one point where the parabola y = x^2-7x+15 and line y = 2x-5 intersect.


The other solution is (x,y) = (5,5)
You'll plug x = 5 into either equation to find that y = 5 pairs up with it. 


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Answers: <font color=red>(4,3) and (5,5)</font>


Visual Verification
{{{drawing(500,500,-3,7,-3,7,
grid(1),
graph(500,500,-3,7,-3,7,0,x^2-7x+15,2x-5),
circle(4,3,0.05),
circle(4,3,0.08),
circle(4,3,0.10),
circle(5,5,0.05),
circle(5,5,0.08),
circle(5,5,0.10),
locate(4,3,"(4,3)"),
locate(5,5,"(5,5)")
)}}}
I recommend using Desmos or GeoGebra as graphing tools to verify the answer. 
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