# Thread: Solving for y when it appears on both sides of equation

1. ## Solving for y when it appears on both sides of equation

I've gotten stuck trying to solve the equation below for y. It's a tricky one for me as my algebra skills are basic at best.

I have managed to get somewhere but the answer that comes out of the final equation is about 17% error from starting equation when I plug some numbers into it.

Have I missed something or is there a better way?

1. $y^2=r^2+x^2$ and
2. $y=\frac{wx}{r}-d$
Need to substitute equation 2 into 1 and solve for y. Then resubstitute and solve for x. (Only r, w, and d are known).

I got to this:

3. $y=\frac{w}{r}(\sqrt{y^2-r^2})-d$

But this leaves me with $y^2$
on the RHS under the square root. From here is were I am unsure.

I managed to get to this:

4. $y=\{\frac{1}{1-(\frac{r}{w})^2}[(\frac{dr}{w})^2+r^2]\}^0^.^5$

This got rid of y and x from RHS however, when I input some values to check if it works I get a different value for y between equations 3 and 4.

Values I used are:

r=3
w=12
d=11
x=4
y=5
(had to use a number for y in eq. 3 to check it works)

Same numbers in eq.4 y = 4.341?

Thanks.

2. ## Re: Solving for y when it appears on both sides of equation

Originally Posted by Cartermgg
I've gotten stuck trying to solve the equation below for y. It's a tricky one for me as my algebra skills are basic at best.

I have managed to get somewhere but the answer that comes out of the final equation is about 17% error from starting equation when I plug some numbers into it.

Have I missed something or is there a better way?

1. $y^2=r^2+x^2$ and
2. $y=\frac{wx}{r}-d$
Need to substitute equation 2 into 1 and solve for y. Then resubstitute and solve for x. (Only r, w, and d are known).

I got to this:

3. $y=\frac{w}{r}(\sqrt{y^2-r^2})-d$

But this leaves me with $y^2$
on the RHS under the square root. From here is were I am unsure.

I managed to get to this:

4. $y=\{\frac{1}{1-(\frac{r}{w})^2}[(\frac{dr}{w})^2+r^2]\}^0^.^5$

This got rid of y and x from RHS however, when I input some values to check if it works I get a different value for y between equations 3 and 4.

Values I used are:

r=3
w=12
d=11
x=4
y=5
(had to use a number for y in eq. 3 to check it works)

Same numbers in eq.4 y = 4.341?

Thanks.
[*] $y^2=r^2+x^2$ and[*] $y=\frac{wx}{r}-d$

$(\frac{wx}{r}-d)^2=r^2+x^2$

Now, expand ans solve for x.

3. ## Re: Solving for y when it appears on both sides of equation

Thank you, I will have a go at it this way.

4. ## Re: Solving for y when it appears on both sides of equation

Originally Posted by Also sprach Zarathustra
[*] $y^2=r^2+x^2$ and[*] $y=\frac{wx}{r}-d$

$(\frac{wx}{r}-d)^2=r^2+x^2$

Now, expand ans solve for x.

I've tried long and hard to solve this equation but I can only get so far. I think I need a bit more guidance.

I have managed to get to this:

$x^2(1-\frac{w^2}{r^2})+2\frac{wxd}{r}=d^2-r^2$

but I can't get any further. How do I get rid of the second x on the LHS?

Thanks

5. ## Re: Solving for y when it appears on both sides of equation

Subtract $d^2-r^2$ from both sides:

$\left(1-\dfrac{w^2}{r^2}\right)x^2 + \left(\dfrac{2wd}{r}\right)x - (d^2-r^2) = 0$

Although it looks complicated this is a quadratic equation in x with
1. $=1 - \dfrac{w^2}{r^2}$
2. $= \dfrac{2wd}{r}$
3. $=d^2-r^2$

I would strongly recommend using the formula to solve this one.

6. ## Re: Solving for y when it appears on both sides of equation

Originally Posted by Cartermgg
1. $y^2=r^2+x^2$ and
2. $y=\frac{wx}{r}-d$
Just divide eq1 by eq2 to get right away:
y = r(r^2 + x^2) / (wx - dr)

7. ## Re: Solving for y when it appears on both sides of equation

Originally Posted by Wilmer
Just divide eq1 by eq2 to get right away:
y = r(r^2 + x^2) / (wx - dr)
That didn't eliminate the $\displaystyle x$... The point of solving the equations simultaneously is to get $\displaystyle x, y$ in terms of the other variables only.

8. ## Re: Solving for y when it appears on both sides of equation

I was answering his opening problem:

"I've gotten stuck trying to solve the equation below for y"