# Thread: Algebra Matrices

1. ## Algebra Matrices

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!

2. No matrices needed, if anything it will just make life harder.

Originally Posted by play60
3x^2 + 4y = 17
2x^2 + 5y = 2
mulitply the first equation through by 5 and the second through by 4, then eliminate y by subtracting.

Originally Posted by play60

xy = 12
y = x - 4

$\displaystyle \displaystyle xy = 12 \implies x(x-4) = 12 \implies x^2-4x-12 = 0 \implies (x=6)(x+2)=0$

3. 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.

4. 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.

5. 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!

6. 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.