# Nonlinear Systems of Equations

• Apr 20th 2008, 12:06 PM
Ballplaya4237
Nonlinear Systems of Equations
Test Tommorow - Just can't seem to get the hang of it...

1) \$\displaystyle y+x^2=4x\$
\$\displaystyle y+4x=16\$

2) \$\displaystyle x^2y=16\$
\$\displaystyle y^2-x^2+16=0\$

3) \$\displaystyle x-2y=2\$
\$\displaystyle y^2-x^2=2x+4\$

Any help is greatly appreciated..thanks
• Apr 20th 2008, 12:13 PM
Moo
Hello,

For 1), isolate y in the second equation and plug it in the first equation.

For 2), isolate y in the first equation, then plug it in the second and solve for X=x² like any quadratic equation

For 3), isolate x in the first and plug in the second.
• Apr 20th 2008, 12:13 PM
topsquark
Quote:

Originally Posted by Ballplaya4237
Test Tommorow - Just can't seem to get the hang of it...

1) \$\displaystyle y+x^2=4x\$
\$\displaystyle y+4x=16\$

2) \$\displaystyle x^2y=16\$
\$\displaystyle y^2-x^2+16=0\$

3) \$\displaystyle x-2y=2\$
\$\displaystyle y^2-x^2=2x+4\$

Any help is greatly appreciated..thanks

For each of these, pick one of the equations and solve for one unknown. Then plug that into the other equation.

For example, the first one I would solve the bottom equation for y:
\$\displaystyle y = -4x + 16\$
and put that into the top equation:
\$\displaystyle (-4x + 16) + x^2 = 4x\$

Solve this for x and use that to solve for y.

-Dan
• Apr 20th 2008, 12:52 PM
Ballplaya4237
Okay, I managed to solve both numbers 1 and 3...

But, 2 still eludes me...

I messed up copying the first time - Here's the correct number 2 equation

2) \$\displaystyle x^2y=16\$
\$\displaystyle x^2+4y+16=0\$
• Apr 20th 2008, 12:55 PM
Moo
This is the same princip : isolate y in the first equation and plug it in the second :)

Multiply the final equation by x², then substitute u=t² and solve it as a quadratic equation ;)