nonlinear problem

[bandaidgirl]

Solve the following nonlinear system by the addition method:

16x^2 - 4y^2 - 72 = 0
. ..x^2 -. .y^2 -. .3 = 0

My work:

I multiplied the second equation by -4, and added down, giving me:

     16x^2 - 4y^2 - 72 = 0
(-4)[  x^2 -  y^2 -  3 = 0]
     
     16x^2 - 4y^2 - 72 = 0
     -4x^2 + 4y^2 + 12 = 0
     ----------------------
     12x^2        - 60 = 0

Then I added the 60 to the other side, and divided both sides by 12:

. . .12x^2 = 60

. . .(12x^2) / 12 = 60 / 12

. . .x^2 = 5

I know both sides have to be squares, but then where do I go from here? Or have I not done the first part right...?

 

[jonboy]

Great job for \(\displaystyle x\)! Now for \(\displaystyle y\):

\(\displaystyle \L \;16x^2\,-\,4y^2\,-\,72\,=\,0\,\Rightarrow\,16x^2\,-\,4y^2\,=\,72\)

\(\displaystyle \L \;-16(x^2\,-\,y^2\,)=(3)-16\,\Rightarrow\,-16x^2\,+\,16y^2\,=\,-48\)


Equate these two final equations: \(\displaystyle \L \;12y^2\,=\,24\)

\(\displaystyle \L \;y^2\,=\,2\)

\(\displaystyle \L \;y\,=\,\sqrt{2}\)

So, as you can see, whenever you wanna to solve for a variable, eradicate the other one. Keep up the great work!

 

[jonboy]

i know both sides have to be squares but then where do i go from here or have i not done the first part right

You mean square rooted.

 

[Denis]

Not quite jonboy.

Simply substitute x^2 = 5 in this equation:

x^2 - y^2 - 3 = 0
5 - y^2 - 3 = 0
y^2 = 2

 

[jonboy]

Not quite jonboy.

Simply substitute x^2 = 5 in this equation:

x^2 - y^2 - 3 = 0
5 - y^2 - 3 = 0
y^2 = 2

Eeek ok just edited my post.