Nonlinear System Elimination Method

**jonb** · November 29th 2009, 11:52 AM

Daughter preparing for test.

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

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

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

$4x^2$ =36
$x^2$ =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

**jonb** · November 29th 2009, 12:09 PM

Ok, now I am stuck.

$3x^2$ -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 $x^2$ in it.

**wonderboy1953** · November 29th 2009, 12:22 PM

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

**wonderboy1953** · November 29th 2009, 12:24 PM

I hope you remember how to solve a quadratic equation.

**jonb** · November 29th 2009, 12:38 PM

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.

**jonb** · November 29th 2009, 12:52 PM

quadratic equation

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

**sammy28** · November 29th 2009, 01:10 PM

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

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

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

subtract to elimate the common term.

$x^2 = 9$

so

$x=\pm 3$

now plug 9 into the equations to solve for y

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

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

$y=\pm 2$

**jonb** · November 29th 2009, 01:16 PM

Ah, same answers....different method

This one is still giving me fits.

-y=11
2x+y=-2

**sammy28** · November 29th 2009, 01:31 PM

hi,

you can use substituion on that one.

$y=-2-2x$

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

so solve the quadratic

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

use the form $ax^2+bx+c=p(x+q)^2+r$ 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.

**jonb** · November 29th 2009, 01:54 PM

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

what if there is nothing that is squared?

X-5y=23
XY=-24

**sammy28** · November 29th 2009, 02:10 PM

hi,

are you sure its not squared?

$x-23=5y$

so

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

so solve

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

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

**jonb** · November 29th 2009, 03:26 PM

Thanks for the help.

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

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