SOLUTION: solve the system
x^2+y^2=100
y=x^2-3x-10
y=x+2
Algebra.Com
Question 1199235: solve the system
x^2+y^2=100
y=x^2-3x-10
y=x+2
Found 2 solutions by ankor@dixie-net.com, greenestamps:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
solve the system
x^2+y^2=100
y=x^2-3x-10
y=x+2
:
y=y therefore
x^2 - 3x - 10 = x + 2
combine on the left
x^2 - 4x - 12 = 0
Factors
(x-6)(x+2) = 0
x = 6, x = -2
:
solve the 1st equation using x=6
6^2 + y^2 = 100
y^2 = 100 - 36
y^2 = 64
y = 8
:
Solve using x=-2
-2^2 + y^2 = 100
y^2 = 100 - 4
y^2 = 96
y =
y ~ 7.8
:
Valid integer solutions x=6, y=8
:
Confirm this in all 3 equations
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
There are three equations and only two variables, so the system might be inconsistent. Pick two of the equations to solve simultaneously and see if the solution satisfies the third equation.
Substitute the third equation in the first:
or
The possible solutions are (x,y)=(-8,-6) and (x,y)=(6,8)
See whether each solution satisfies the second given equation.
x = -8: x^2-3x-10 = 64+24-10 = 78 NO
x = 6: x^2-3x-10 = 36-18-10 = 8 YES
ANSWER: x=6; y=8
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