SOLUTION: How would you solve the quadratic linear system algebraically which is y=4x^2-6x-10 and y=25-2x?

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: How would you solve the quadratic linear system algebraically which is y=4x^2-6x-10 and y=25-2x?      Log On


   

Question 1005383: How would you solve the quadratic linear system algebraically which is y=4x^2-6x-10 and y=25-2x?
Found 2 solutions by josgarithmetic, macston:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Substitution method; solve either equation if necessary, for either variable and substitute into the other equation. Your exercise gives two formulas for y, so they can simply be equated, and solve for x.

You would eventually find for x using general solution for quadratic equation,
x=%284%2B-+24%29%2F4
x=%281%2B-+6%29
x=-5 or x=7
Find the y which correspond to each of those.

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
y=4x%5E2-6x-10
y=25-2x
Since both equal y:
4x%5E2-6x-10=25-2x
4x%5E2-4x-35=0
%284x%2B10%29%28x-3.5%29=0
4x%2B10=0 OR x-3.5=0
4x=-10 OR x=3.5
x=-2.5 OR x=3.5
.
Using x=(-2.5):
y=25-2x
y=25-2%28-2.5%29
y=25-%28-5%29
y=30
.
CHECK (Using other equation):
y=4x%5E2-6x-10
30=4%28-2.5%29%5E2-6%28-2.5%29-10
30=4%286.25%29-%28-15%29-10
30=25%2B15-10
30=40-10
30=30
.
Using x=3.5:
y=25-2x
y=25-2%283.5%29
y=25-7
y=18
.
CHECK:
y=4x%5E2-6x-10
18=4%283.5%29%5E2-6%283.5%29-10
18=4%2812.25%29-21-10
18=49-31
18=18
.
ANSWER: The solutions are (-2.5,30) and (3.5,18)