SOLUTION: solve the sytem algebraically y^2 = x^2 - 32 x^2 + y^2 = 40

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: solve the sytem algebraically y^2 = x^2 - 32 x^2 + y^2 = 40      Log On


   

Question 76853: solve the sytem algebraically
y^2 = x^2 - 32
x^2 + y^2 = 40
Found 2 solutions by checkley75, funmath:
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
Y^2=X^2-32 NOW SUBSTITUTE (X^2-32) FOR Y^2 IN THE SECOND EQUATION & SOLVE FOR X
X^2+(X^2-32)=40
X^2+X^2=40+32
2X^2=72
X^2=72/2
X^2=36
X=6 ANSWER.
Y*2=6*6-32
Y^2=36-32
Y^2=4
Y=2 ANSWER.
PROOF
6^2+2^2=40
36+4=40
40=40

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
solve the sytem algebraically
y^2 = x^2 - 32
x^2 + y^2 = 40
Substitute x^2-32 in for y^2 in the second equation.
x%5E2%2B%28x%5E2-32%29=40
2x%5E2-32=40
2x%5E2-32%2B32=40%2B32
2x%5E2=72
2x%5E2%2F2=72%2F2
x%5E2=36
sqrt%28x%5E2%29=%2B-sqrt%2836%29
x=-6 or x=6
Let x=-6 or x=6 (They will both have the same outcome because x is squared.) in the first equation and solve for y.
y%5E2=%28-6%29%5E2-32
y%5E2=36-32
y%5E2=4
sqrt%28y%5E2%29=%2B-sqrt%284%29
y=-2 or y=2
Therefore two solutions are (-6,-2) and (-6,2)
y%5E2=%286%29%5E2-32
y%5E2=36-32
y%5E2=4
sqrt%28y%5E2%29=%2B-sqrt%284%29
y=-2 or y=2
Therefore two solutions are (6,-2) and (6,2)
There are four solutions: (-6,-2), (-6,2), (6,-2), and (6,2)
It helps to look at the graph.
Graphically it looks like this:

Happy Calculating!!!