# Thread: Nonlinear Systems

1. ## Nonlinear Systems

Solve by substitution.

$\displaystyle xy=3$

$\displaystyle x^2+y^2=10$

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.

2. 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

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

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

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

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

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

3. Originally Posted by HSS8
Solve by substitution.

$\displaystyle xy=3$

$\displaystyle x^2+y^2=10$

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

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

so the second

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

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

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

Edit. Too late.

4. Originally Posted by HSS8
Solve by substitution.

$\displaystyle xy=3$

$\displaystyle x^2+y^2=10$

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 $\displaystyle x \ne 0$ then

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

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

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

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

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

Plugging into the first equaion you get the four solutions

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

Edit I am the latest of all

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

But anyway, the solution is simple.

Solve for either x or y and then substitute.

$\displaystyle xy=3$

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

Now substituting and solving for x

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

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

mutiplying bot sides by $\displaystyle x^2$

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

subtracting

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

Factoring

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

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

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

So.....

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

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

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

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

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