SOLUTION: Choose the real solutions of the system: x = (y^2) + -9 and x - 4y - 12 = 0 I tried using the substitution method but I keep getting stuck!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Choose the real solutions of the system: x = (y^2) + -9 and x - 4y - 12 = 0 I tried using the substitution method but I keep getting stuck!      Log On


   

Question 757031: Choose the real solutions of the system:
x = (y^2) + -9 and x - 4y - 12 = 0
I tried using the substitution method but I keep getting stuck!
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
x = (y^2) + -9 and x - 4y - 12 = 0
x=4y%2B12 from second equation. Substitute into the first equation.
4y%2B12=y%5E2-9
y%5E2-4y-9-12=0
y%5E2-4y-21=0
You should no longer be stuck. The lefthand side is factorable.




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Just a little bit of review:
%28x%2Bh%29%28x%2Bk%29=x%5E2%2B%28h%2Bk%29x%2Bhk

If you find some square trinomial as any general x%5E2%2Bax%2Bb and you want to factor it, try thinking of it as x%5E2%2Bax%2Bb=x%5E2%2B%28h%2Bk%29x%2Bhk
So you are looking for two numbers, h and k, so that hk=b and h+k=a.
What you do is look for all the two-number factorizations for b, and find which ones give you a sum of a. We then test each combination until we find something that works. There is a somewhat more analytical way also, but the search and test method is usually enough.