SOLUTION: Determine the coordinates of the point(s) of intersection for the following linear-quadratic system algebraically: y = x^2-7x+15 and y = 2x-5.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Determine the coordinates of the point(s) of intersection for the following linear-quadratic system algebraically: y = x^2-7x+15 and y = 2x-5.      Log On


   

Question 1197265: Determine the coordinates of the point(s) of intersection for the following linear-quadratic system algebraically:
y = x^2-7x+15 and y = 2x-5.
Found 4 solutions by ewatrrr, ikleyn, josgarithmetic, math_tutor2020:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
Determine the coordinates of the point(s) of intersectio
y = x^2-7x+15 and y = 2x-5.
 x^2-7x+15  = 2x-5
 x^2 - 9x + 20 = 0   | Factor
(x - 4)(x - 5) = 0 
  x = 4  0r x = 5
And
y = 3  0r y = 5      |  Substitute value of x into y = 2x-5
point(s) of intersection:  P(4,3)and P(5,5)
Graphing 
x^2-7x+15  Green
y = 2x-5   Blue

Wish You the Best in your Studies.


Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.

Reduce two starting equations to one single equation for x, eliminating y

    x^2 - 7x + 15 = 2x - 5


Write it in the standard form

    x^2 - 9x + 20 = 0


Factor left side

    (x-4)*(x-5) = 0


The roots are 4 and 5.


For each root calculate y


    y%5B1%5D = 2%2Ax%5B1%5D-5 = 2*4 - 5 = 3;

    y%5B2%5D = 2%2Ax%5B2%5D-5 = 2*5 - 5 = 5.


ANSWER.  There are 2 solutions:  (1)  (x,y) = (4,3)  and  (2)  (x,y) = (5,5).

Solved, with full explanations.



Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
substituting for y
x%5E2-7x%2B15=2x-5
x%5E2-9x%2B20=0
factorize the quadratic member
%28x-4%29%28x-5%29=0
x values for the intersection at 4, and 5.


graph%28400%2C400%2C-2%2C10%2C-2%2C10%2Cx%5E2-7x%2B15%2C2x-5%29

Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

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+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28-9%29%2B-sqrt%28%28-9%29%5E2-4%281%29%2820%29%29%29%2F%282%281%29%29

x+=+%289%2B-sqrt%2881+-+80%29%29%2F%282%29

x+=+%289%2B-sqrt%281%29%29%2F%282%29

x+=+%289%2B-++1%29%2F%282%29

x+=+%289%2B1%29%2F%282%29 or x+=+%289-1%29%2F%282%29

x+=+%2810%29%2F%282%29 or x+=+%288%29%2F%282%29

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: (4,3) and (5,5)

Visual Verification

I recommend using Desmos or GeoGebra as graphing tools to verify the answer.