# Thread: [SOLVED] Solving simple Nonlinear equation ?

1. ## [SOLVED] Solving simple Nonlinear equation ?

So i need to find the X and Y values ..

xy = 3
x^2 + y^2 = 10

So i solved the first equation for y = 3/x
Plugged in the second equation, but can't solve for x !
x^2 + (3/x)^2 = 10

2. All that you've done so far is correct.
You need to sub y = 3/x into either of the original equations, so you will get

x^2 * 3/x = 1
or
x^2 + 9/x^2 = 10
the first equation is probably simpler

you then simplify it to get 3x^2/x = 3
then multiply both sides by x so that
3x^2=3x
and so you can see that x = 1

i hope that helped=]

3. The X has 4 possible values 3, -3, 1. -1
How exactly should i proceed to find all the values of X and back substitute to find the Y's

4. Originally Posted by xterminal01
So i need to find the X and Y values ..

xy = 3
x^2 + y^2 = 10

So i solved the first equation for y = 3/x
Plugged in the second equation, but can't solve for x !
x^2 + (3/x)^2 = 10
Originally Posted by xterminal01
The X has 4 possible values 3, -3, 1. -1

How exactly should i proceed to find all the values of X and back substitute to find the Y's
$x^2 + \frac{9}{x^2} = 10$

Multiply both sides by $x^2$:

$\Rightarrow x^4 + 9 = 10 x^2 \Rightarrow x^4 - 10 x^2 + 9 = 0 \Rightarrow (x^2 - 1) (x^2 - 9) = 0$.

Therefore either $x^2 - 1 = 0$ or $x^2 - 9 = 0$. I'm sure you can solve these two ....

Originally Posted by ella85
All that you've done so far is correct.
You need to sub y = 3/x into either of the original equations, so you will get

x^2 * 3/x = 1 Mr F says: What equation have you used here?

or x^2 + 9/x^2 = 10

the first equation is probably simpler Mr F says: The first equation is xy = 3 => y = 3/x. Which is where the OP got y = 3/x from in the first place.

you then simplify it to get 3x^2/x = 3 Mr F says: The first equation is NOT x^2 y = 3 .....

then multiply both sides by x so that
3x^2=3x
and so you can see that x = 1

i hope that helped=]
Sorry but none of this makes any sense.

5. Hello, xterminal01!

Solve the system: . $\begin{array}{cc}xy \:=\: 3 & {\color{blue}[1]} \\ x^2 + y^2 \:=\: 10 & {\color{blue}[2]}\end{array}$
If you want to avoid fractions, here is an imaginative approach . . .

. . $\begin{array}{ccccccc}\text{Multiply {\color{blue}[1]} by 2:} & 2xy &=& 6 \\ \text{Add {\color{blue}[2]}:} & x^2+y^2&=&10 \end{array}$

. . And we have: . $x^2+2xy + y^2 \:=\:16 \quad\Rightarrow\quad (x+y)^2 \:=\:16$

Hence: . $x + y \:=\:\pm4 \quad\Rightarrow\quad y \;=\;-x \pm4$

Substitute into [1]: . $x[-x\pm4] \:=\:3$

. . and solve the resulting two quadratics.

6. Thanks i was able to solve everything perfectly fine.
One question, when you added x^2 to the 10 and added x^2 on the left side to clear the fraction.. Why did you as well add to x^2 before it became x^4.. Wouldn't that mean that you added on the left side two x^2 and on the right side just one"? Doesn't that mean that the equation is unbalanced now? What you do to one side you have to do exactly to the other one as well ?

Originally Posted by mr fantastic
$x^2 + \frac{9}{x^2} = 10$

Multiply both sides by $x^2$:

$\Rightarrow x^4 + 9 = 10 x^2 \Rightarrow x^4 - 10 x^2 + 9 = 0 \Rightarrow (x^2 - 1) (x^2 - 9) = 0$.

Therefore either $x^2 - 1 = 0$ or $x^2 - 9 = 0$. I'm sure you can solve these two ....

Sorry but none of this makes any sense.

7. Originally Posted by xterminal01
Thanks i was able to solve everything perfectly fine.
One question, when you added x^2 to the 10 and added x^2 on the left side to clear the fraction.. Why did you as well add to x^2 before it became x^4.. Wouldn't that mean that you added on the left side two x^2 and on the right side just one"? Doesn't that mean that the equation is unbalanced now? What you do to one side you have to do exactly to the other one as well ?
Originally Posted by mr fantastic
$x^2 + \frac{9}{x^2} = 10$

Multiply both sides by ${\color{red}x^2}$:

[snip]
..

8. x^2 and 9/x^2 was multiplied, thus both terms on the left side were multiplied with x^2?
My question is why isn't x^2-10x^2 + 9 = 0 the result of the multiplication

Originally Posted by mr fantastic
..

9. Originally Posted by xterminal01
x^2 and 9/x^2 was multiplied, thus both terms on the left side were multiplied with x^2?
My question is why isn't x^2-10x^2 + 9 = 0 the result of the multiplication
$x^2 + \frac{9}{x^2} = 10$

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

Do you know how to expand the left hand side?

10. I did not realize that you are suposse to multiply the whole left side.
So powers would add up and fraction would clear.
Thanks for you help

Originally Posted by mr fantastic
$x^2 + \frac{9}{x^2} = 10$

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

Do you know how to expand the left hand side?