Nonlinear Systems

**HSS8** · May 17th 2009, 01:13 PM

Solve by substitution.

$xy=3$

$x^2+y^2=10$

I have a linear equation and a circle, right? I tried isolating either x or y in equation 1 and substituting into equation 2 then back substituting the answers, but didn't get the right answer.

**derfleurer** · May 17th 2009, 01:19 PM

My algebra might be rough but xy=3 is clearly not linear. =p

Now if xy = 3, than y = 3/x

So sub for y

$x^2 + (\frac 3x)^2 = 10$

$x^2 + \frac 9{x^2} = 10$

$x^4 + 9 = 10x^2$

$x^4 - 10x^2 + 9 = 0$

$(x^2 - 1)(x^2 - 9) = 0$

**Danny** · May 17th 2009, 01:20 PM

Originally Posted by HSS8

Solve by substitution.

$xy=3$

$x^2+y^2=10$

I have a linear equation and a circle, right? I tried isolating either x or y in equation 1 and substituting into equation 2 then back substituting the answers, but didn't get the right answer.

Actually both are nonlinear. I think it's easier to solve the first for y

$ y = \frac{1}{x} $

so the second

$ x^2 + \frac{9}{x^2} = 10 $
or

$ x^4 - 10x^2 + 9 = 0 $

which factors $(x^2-9)(x^2-1) = (x-3)(x+3)(x-1)(x+1) = 0$

Edit. Too late.

**TheEmptySet** · May 17th 2009, 01:23 PM

Originally Posted by HSS8

Solve by substitution.

$xy=3$

$x^2+y^2=10$

I have a linear equation and a circle, right? I tried isolating either x or y in equation 1 and substituting into equation 2 then back substituting the answers, but didn't get the right answer.

Suppose that $x \ne 0$ then

$y=\frac{3}{x}$ and sub this into the other equation to get

$x^2+\left( \frac{3}{x}\right)^2=10 \iff x^2+\frac{9}{x^2}=10$

Now if we multiply by $x^2$ we get (it is quadratic in form)

$(x^2)^2+9=10x^2 \iff (x^2)^2-10x^2+9=0 \iff (x^2-9)(x^2-1)=0$

$(x-3)(x+3)(x-1)(x+1)=0$

Plugging into the first equaion you get the four solutions

$(x,1);(-3,-1);(1,3);(-1,-3)$

Edit I am the latest of all

**VonNemo19** · May 17th 2009, 01:43 PM

First of all, you don't have a linear equation. The equation $xy=3$ is a relative of the inverse function $y=\frac{1}{x}$ with each element in the range of the function increased by a factor of $3$ .

But anyway, the solution is simple.

Solve for either x or y and then substitute.

$xy=3$

$y=\frac{3}{x}$

Now substituting and solving for x

$x^2+(\frac{3}{x})^2=10$

$x^2+\frac{9}{x^2}=10$

mutiplying bot sides by $x^2$

$x^4+9=10x^2$

subtracting

$x^4-10x^2+9=0$

Factoring

$(x^2-9)(x^2-1)=0$
and this implies that

$x=\pm{\sqrt{9}}=\pm{3},\pm{\sqrt{1}}=\pm{1}$

Now we have to find all solutions to the equation $y=\frac{3}{x}$ when
$x=\pm{\sqrt{9}}=\pm{3},\pm{\sqrt{1}}=\pm{1}$

So.....

$y=\frac{3}{+3}=1$

$y=\frac{3}{-3}=-1$

$y=\frac{3}{+1}=3$

$y=\frac{3}{-1}=-3$

So, the points of importance are $(3,1),(-3,-1),(1,3),(-1,-3)$

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