Nonlinear Systems solved by Elimination

[flappergirl68]

I've tried several different ways to solve this problem by elimination and just can't get it right, any suggestions??

2x+y^2=36

x+y=14


Thanks!

 

[o_O]

What have you done so far?

 

[stapel]

...any suggestions??

2x+y^2=36
x+y=14

There are many ways to proceed with this: graphing, substitution and then completing the square, etc, etc.

It would probably be best to use the method with which you're most comfortable. So what are you trying, and how far have you gotten?

Please be complete. Thank you!

Eliz.

 

[Mrspi]

I've tried several different ways to solve this problem by elimination and just can't get it right, any suggestions??

2x+y^2=36

x+y=14


Thanks!

You could multiply both sides of the second equation by -2...

-2x - 2y = -28

Now, add this to the first equation, and you'll eliminate x. The resulting quadratic equation in y can be solved by your favorite method.

 

[flappergirl68]

I had multiplied the second equation by -2, this was my first inkling, but had trouble factoring the resulting quadratic equation which is

y^2-2y+28=0

and ran across the same sort or problem when I squared everything in the second part and multiplied by -1.

 

[stapel]

I had...trouble factoring the resulting quadratic equation which is y^2-2y+28=0

How did you arrive at this equation? You had:

. . . . .2x + y² = 36

. . . . .2x + 2y = 28

...and then you subtracted, making the "x" terms cancel off. But where did the "28" in your final equation come from?

Please be complete. Thank you!

Eliz.

 

[Mrspi]

After you multiply both sides of the second equation by -2 (as I showed in an earlier response), your system looks like this:

2x + y^2 = 36
-2x - 2y = -28

ADD the two equations together:

2x + y^2 = 36
-2x - 2y = -28
-----------------

y^2 - 2y = 8

Now...proceed from there.