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
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=]
$\displaystyle x^2 + \frac{9}{x^2} = 10$
Multiply both sides by $\displaystyle x^2$:
$\displaystyle \Rightarrow x^4 + 9 = 10 x^2 \Rightarrow x^4 - 10 x^2 + 9 = 0 \Rightarrow (x^2 - 1) (x^2 - 9) = 0$.
Therefore either $\displaystyle x^2 - 1 = 0$ or $\displaystyle x^2 - 9 = 0$. I'm sure you can solve these two ....
Sorry but none of this makes any sense.
Hello, xterminal01!
If you want to avoid fractions, here is an imaginative approach . . .Solve the system: .$\displaystyle \begin{array}{cc}xy \:=\: 3 & {\color{blue}[1]} \\ x^2 + y^2 \:=\: 10 & {\color{blue}[2]}\end{array}$
. . $\displaystyle \begin{array}{ccccccc}\text{Multiply {\color{blue}[1]} by 2:} & 2xy &=& 6 \\ \text{Add {\color{blue}[2]}:} & x^2+y^2&=&10 \end{array}$
. . And we have: .$\displaystyle x^2+2xy + y^2 \:=\:16 \quad\Rightarrow\quad (x+y)^2 \:=\:16 $
Hence: .$\displaystyle x + y \:=\:\pm4 \quad\Rightarrow\quad y \;=\;-x \pm4$
Substitute into [1]: .$\displaystyle x[-x\pm4] \:=\:3$
. . and solve the resulting two quadratics.
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 ?