Can some one show me how to get the final expression?

[nasi112]

$$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$
$$\frac{1}{x + f} + \frac{1}{y + f} = \frac{1}{f}$$
$$xy = f^2$$

 

[fresh_42]

Multiply the second equation with $(x+f)\cdot (y+f) \cdot f.$

 

[nasi112]

Multiply the second equation with $(x+f)\cdot (y+f) \cdot f.$

Thanks. Still can't get the expression.

 

[fresh_42]

Thanks. Still can't get the expression.

After the multiplication, you need to cancel the remaining quotients, multiply out the numerators using the distributive law $a\cdot (b+c)=a\cdot b+a\cdot c$, and subtract what occurs on both sides.

Where exactly are you stuck?

 

[nasi112]

After the multiplication, you need to cancel the remaining quotients, multiply out the numerators using the distributive law $a\cdot (b+c)=a\cdot b+a\cdot c$, and subtract what occurs on both sides.

Where exactly are you stuck?

The left hand side will only cancel $$x + f$$ how to get rid of $$y + f$$?

 

[nasi112]

After the multiplication, you need to cancel the remaining quotients, multiply out the numerators using the distributive law $a\cdot (b+c)=a\cdot b+a\cdot c$, and subtract what occurs on both sides.

Where exactly are you stuck?

Thanks. Got it. I had to distribute.

 

[fresh_42]

We have

$$\dfrac{1}{f}=\dfrac{1}{x+f}+\dfrac{1}{y+f}$$
and multiplication with $f\cdot (x+f)\cdot (y+f)$ gives us

$$\dfrac{f\cdot (x+f)\cdot (y+f)}{f}=\dfrac{f\cdot (x+f)\cdot (y+f)}{x+f}+\dfrac{f\cdot (x+f)\cdot (y+f)}{y+f} .$$
Now, each denominator occurs in the numerator, too. That's why we may cancel them and get

$$\dfrac{ (x+f)\cdot (y+f)}{1}=\dfrac{f\cdot (y+f)}{1}+\dfrac{f\cdot (x+f)}{1}$$or simply

$$(x+f)\cdot (y+f)=f\cdot (y+f)+f\cdot (x+f).$$
Can you go on from here?

 

[fresh_42]

Thanks. Got it. I had to distribute.

Well, that is everything we have that connects addition and multiplication!

 

[nasi112]

We have

$$\dfrac{1}{f}=\dfrac{1}{x+f}+\dfrac{1}{y+f}$$
and multiplication with $f\cdot (x+f)\cdot (y+f)$ gives us

$$\dfrac{f\cdot (x+f)\cdot (y+f)}{f}=\dfrac{f\cdot (x+f)\cdot (y+f)}{x+f}+\dfrac{f\cdot (x+f)\cdot (y+f)}{y+f} .$$
Now, each denominator occurs in the numerator, too. That's why we may cancel them and get

$$\dfrac{ (x+f)\cdot (y+f)}{1}=\dfrac{f\cdot (y+f)}{1}+\dfrac{f\cdot (x+f)}{1}$$or simply

$$(x+f)\cdot (y+f)=f\cdot (y+f)+f\cdot (x+f).$$
Can you go on from here?

My mistake I did $$(x+f)\cdot (y+f) \cdot f \frac{1}{x + f} + \frac{1}{y + f} = (x+f)\cdot (y+f) \cdot f \frac{1}{f}$$
Got it. Thanks.

 

[The Highlander]

The left hand side will only cancel $$x + f$$ how to get rid of $$y + f$$?

Don't (try to) "get rid" of it too soon.

I would suggest this...

$$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$
$$\implies\frac{d_i}{d_od_i} + \frac{d_o}{d_od_i} = \frac{1}{f}$$
$$\implies\frac{d_i+d_o}{d_od_i} = \frac{1}{f}$$
$$\implies d_i+d_o = \frac{d_od_i}{f}$$
$$\implies f(d_i+d_o) = d_od_i$$
$$\text{but, since } d_o = (x+f) \text{ and } d_i=(y+f ) \text{, then...}\\d_i+d_o=(x+y+2f)\text{ and }d_od_i=(x+f)(y+f)=xy+xf+fy+f^2$$
$$\implies f(x+y+2f) = xy+xf+fy+f^2$$
$$\implies fx+fy+2f^2 = xy+xf+fy+f^2$$
I and it's simply a matter of cancelling from there...

Hope that helps.

 

[The Highlander]

Ah, you've been busy while I was typing, lmao.

 

[khansaheb]

Thanks. Still can't get the expression.

@Nasi - In your previous post (wayyyyy back when) you were trying to solve PDE - and - you cannot do this assignment ??!! This is is beginning algebra - you should be "master" of these !!

 

[Steven G]

Requiring differential equation student to recall beginning algebra is asking for a lot (unfortunately)

 

[The Highlander]

Requiring differential equation student to recall beginning algebra is asking for a lot (unfortunately)

I knew you were just hiding in the corner!!
(On the stool with your name on it? )

"I have been on this forum for just over 10 years and have decided to not be a helper anymore as long as this logistic_guy (he seems to have lost interest, lol ) is still around. I enjoyed my time here, but as I said it's time to say goodbye.
Steven"

Welcome back.

 

[nasi112]

Don't (try to) "get rid" of it too soon.

I would suggest this...

$$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$
$$\implies\frac{d_i}{d_od_i} + \frac{d_o}{d_od_i} = \frac{1}{f}$$
$$\implies\frac{d_i+d_o}{d_od_i} = \frac{1}{f}$$
$$\implies d_i+d_o = \frac{d_od_i}{f}$$
$$\implies f(d_i+d_o) = d_od_i$$
$$\text{but, since } d_o = (x+f) \text{ and } d_i=(y+f ) \text{, then...}\\d_i+d_o=(x+y+2f)\text{ and }d_od_i=(x+f)(y+f)=xy+xf+fy+f^2$$
$$\implies f(x+y+2f) = xy+xf+fy+f^2$$
$$\implies fx+fy+2f^2 = xy+xf+fy+f^2$$
I and it's simply a matter of cancelling from there...

Hope that helps.

Sorry. Just saw your answer. They posted unrelated stuff and that got me lost track of the thread.

 

[The Highlander]

Sorry. Just saw your answer. They posted unrelated stuff and that got me lost track of the thread.

No problem. YVW.