# nonlinear system of equations problem

• May 15th 2013, 08:56 AM
cdbowman42
nonlinear system of equations problem
I can't figure out where I'm going wrong in solving this system of equations.

x2+2xy-y2=14
x2-y2= -16

Elimination

x2+2xy-y2=14
-x2-0xy+y2= 16

2xy=30

Substitution

x2+30-y2=14

x2-y2+30=14

-16+30=14
14=14

I keep getting results like this any way I try to substitute or rearrange the equations.
• May 15th 2013, 09:13 AM
ebaines
Re: nonlinear system of equations problem
Whenever you get an expression like 14=14 that means yuo have essentially made the same substitution twice, so you no longer have two independent equations.

Try this: note that x^2 + 2xy + y^2 = (x+y)^2, and that x^2-y^2 is the difference of two squares. So the two equations become:

$(x-y)^2 = 14$
$(x+y)(x-y) = -16$

Can you take it from here?
• May 15th 2013, 09:40 AM
cdbowman42
Re: nonlinear system of equations problem
The first equation was x^2+2xy-y^2=14 (not x^2 + 2xy + y^2 = 16)

I don't think x^2+2xy-y^2 can be factored any further. Can it?
• May 15th 2013, 10:56 AM
ebaines
Re: nonlinear system of equations problem
Quote:

Originally Posted by cdbowman42
The first equation was x^2+2xy-y^2=14 (not x^2 + 2xy + y^2 = 16)

I don't think x^2+2xy-y^2 can be factored any further. Can it?

You're right - my mistake. I'm afraid I misread it.

OK, try this then: sub the second equation into the first:

$x^2 + 2xy - y^2 = 14 \to (x^2-y^2) + 2xy = 14 \to -16 + 2xy = 14 \to xy = 15$

So we have $xy=15$ and $y = \frac {15} x$. Sub these values back into the first equation:

$x^2 + 30 - \frac {15^2}{x^2} = 14$

$x^4 + 30 x^2 - 225 = 14 x^2$

$x^4 + 16x^2 - 225 = 0$

Let w = x^2, to give a quadratic equation that you can factor:

$w^2 + 16w -225 = (w-9)(w+25) = 0$

Hence w = 9 or -25, and x = +/- 3 and +/- 5i. The corresponding values for y are +/- 5 and -/+ 3i.
• May 15th 2013, 10:58 AM
Soroban
Re: nonlinear system of equations problem
Hello, cdbowman42!

Quote:

$\begin{array}{ccc}x^2 + 2xy - y^2 \:=\:14 & [1] \\ x^2-y^2 \:=\:\text{-}16 & [2] \end{array}$

Subtract [1] - [2]: . $2xy \:=\:30 \quad\Rightarrow\quad y \:=\:\tfrac{15}{x}\;\;[3]$

Substitute into [2]: . $x^2 - \left(\tfrac{15}{x}\right)^2 \:=\:\text{-}16 \quad\Rightarrow\quad x^2 - \tfrac{225}{x^2} \:=\:\text{-}16$

Multiply by $x^2\!:\;\;x^4 - 225 \:=\:\text{-}16x^2 \quad\Rightarrow\quad x^4 + 16x^2 - 225 \:=\:0$

Factor: . $(x^2+25)(x^2-9) \:=\:0$

We have: . $\begin{Bmatrix}x^2 + 25 \:=\:0 & \Rightarrow& x^2 \:=\:\text{-}25 &\Rightarrow& \text{no real roots} \\ x^2 - 9 \:=\:0 & \Rightarrow & x^2 \:=\:9 & \Rightarrow& x \:=\:\pm3 \end{Bmatrix}$

Substitute into [3]: . $y \:=\:\tfrac{15}{\pm3} \:=\:\pm5$

Therefore: . $(x,y) \;=\;(3,5),\;(\text{-}3,\text{-}5)$