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Melody  11 Feb 2022
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bob:

Solve the system:
y = 2x2 - 4x - 4
y = 2x^2 - 4x - 4 OR y = 2x2 - 4x - 4
x + y = -5
I wrote my second formula the same as you Bob but then I highlighted the 2 and pressed the 'sup' button above the smilies. (sup stands for superscript).

the choices are
a. {-6, 1}
b. no real solution
c. {(1, -6) and (0.5,-5.5)}
d. [{1,6}, {-1,-6}]



Thank you for your solution 1<3 Algebra
Maybe it is correct but I think that Bob intended what I wrote above.
Method 1
Now you can do this like 1<3 Algebra did, and if you do that Algebra's (1<3 is too hard to write every time) way (but with the correct original y formulas) you will get to the correct answer.
It does involve solving a quadratic equation. If you need help solving the quadratic please post again.
Method 2
The other method is by elimination.
You know that x+y=-5
look at what you have -6+1=5 good, 1+-6=-5 good, 0.5+-5,5=-5 good, 1+6 doesn't equal -5 so d is definitely wrong.
So far it could be a,b,or c.
-------------------------------------------------
now the choices are
a. {-6, 1}
b. no real solution
c. {(1, -6) and (0.5,-5.5)}
--------------------------------------------------
You also know that
y = 2x 2 - 4x - 4
I'd change it to f(x) = 2x 2 - 4x - 4 just because the notation is easier for me, but you don't have to.
Now I'll test f(-6) if i get an answer of 1 then I know that (a) is a correct answer otherwise it is not a correct answer. That's part (a)
now for part d) I'd test f(1) and f(0.5) to see if the given points work.
Now you either have a or c or nothing.
If neither a nor b are correct then you still can't say for sure that b is the answer. I didn't think of that before I started writing all this.
So maybe you should do it the first way.
10 Jan 2014