Find all solutions to the system of equations.
1. 3x^2 + 4y = 17
2x^2 + 5y = 2
2. xy = 12
y = x - 4
Thanks in advance!
No matrices needed, if anything it will just make life harder.
mulitply the first equation through by 5 and the second through by 4, then eliminate y by subtracting.
$\displaystyle \displaystyle xy = 12 \implies x(x-4) = 12 \implies x^2-4x-12 = 0 \implies (x=6)(x+2)=0$
I apologize for posting more than 2 questions.
I am working on problem 1.
I multiplied the first equation by 5 and the second by 4. I now have:
15 20 85
8 20 8
Is this correct thus far? If so, what is the next step. I know I am to subtract, but am unsure how to do so. Thanks in advance.
What? What happened to the variables, x and y? If you are trying to write these as matrices, that might not work- these are NOT linear equations.
(Actually it will if you think of these as "linear" in $\displaystyle x^2$ and y, not x and y.)
I think you mean
$\displaystyle 15x^2+ 20y= 85$ and
$\displaystyle 8x^2+ 20y= 8$
Subtract the second equation from the first to get a single equation in x.
Thank you. Yes, that is what I intended.
The answer to problem 1 (that I got) is: x² = 11 and y = -4.
The answer to problem 3 (that I got) is: x = -1, y = 0, z = 1
I am struggling with problem 2. How do I set this problem up?
For problem 4, do I use Gaussian elimination or Gauss-Jordan elimination?
Thanks again!
Once again, these are NOT linear equations- "Gaussian Elimination might not work. For #2, the simplest thing to do is to use the second equation, y= x- 4 to replace the "y" in the first equation:
xy= x(x- 4)= 12. That's a quadratic equation for x.
I see no third or fourth equations. I suspect a moderator has removed them since more than two questions in a post is against the rules. If you have more questions, open a new thread.