# Thread: Nonlinear System Elimination Method

1. ## Nonlinear System Elimination Method

Daughter preparing for test.

$\displaystyle 4x^2$+$\displaystyle 3y^2$=48
$\displaystyle 3x^2$+$\displaystyle 2y^2$=35

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$\displaystyle -12x^2$+$\displaystyle -9y^2$=-144
$\displaystyle 12x^2$+$\displaystyle +8y^2$=140
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$\displaystyle -y^2$=-4
$\displaystyle y^2$=4
y=+-2

$\displaystyle 4x^2$+$\displaystyle 3(2)^2$=48

$\displaystyle 4x^2$=36
$\displaystyle x^2$=9
x=+-3

(I think I answered the first part of the question while typing it.)

I know I will have a question in a moment

2. Ok, now I am stuck.

$\displaystyle 3x^2$-y=11
2x+y=-2

Dang this has been more than 20 years ago for me.

how do I proceed with this one. The second part does not have an $\displaystyle x^2$ in it.

3. ## Solution

Solve for y in the second equation and substitute into the first equation and the solution will shake out.

4. ## Followup thought

I hope you remember how to solve a quadratic equation.

5. 2x+y=-2

2x+2=-y

Ok...I am dumb. I just follow the pattern in her study guide. How do I solve for Y?

Even if I get to where I know what is going on then I have to explain it to her. She works as a nurses aid full time and wonders what she will do with this.

I will review that...but not sure how to plug that into this.

7. Hi there,

i thought you were supposed to solve by elimination?

you can elimate either x or y by using an appropriate factor.

multiplying equation 2 by 3 and equation 1 by 2 gives you

$\displaystyle 9x^2+6y^2=105$

$\displaystyle 8x^2+6y^2=96$

subtract to elimate the common term.

$\displaystyle x^2 = 9$

so

$\displaystyle x=\pm 3$

now plug 9 into the equations to solve for y

$\displaystyle 8(9) + 6y^2 = 96$

$\displaystyle y^2 = \frac{96-72}{6} = 4$

$\displaystyle y=\pm 2$

8. Ah, same answers....different method

This one is still giving me fits.

-y=11
2x+y=-2

9. hi,

you can use substituion on that one.

$\displaystyle y=-2-2x$

$\displaystyle 3x^2-(-2-2x)=11$

so solve the quadratic

$\displaystyle 3x^2+2x-9=0$

use the form $\displaystyle ax^2+bx+c=p(x+q)^2+r$ or the formula to solve as discriminant is > 0 but not a perfect square. ie solution has two real roots but cannot be factored to integers.

10. Ok...I can see it when someone else does it. Thanks

what if there is nothing that is squared?

X-5y=23
XY=-24

11. hi,

are you sure its not squared?

$\displaystyle x-23=5y$

so

$\displaystyle x\frac{(x-23)}{5}=-24$

so solve

$\displaystyle x^2-23x+120=0$

dont forget to solve for the $\displaystyle y$ values in this and previous case. The discriminant on this one is > 0 and is a perfect square so will factor to integers.

12. Thanks for the help.

I think I was able to do the next 6 on my own. I think