# Homework Help: Find Y when AxY=B, BxY=C, and B is unknown

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1. Sep 28, 2017

### 1plus1is10

How do I:
Find Y when AxY=B, BxY=C, and B is unknown?
(A and C are known)

Example: If A=100 and C=169 then Y=1.3 and B=130

I assume I use log or pow but cannot figure it out.
Thanks.

2. Sep 28, 2017

### Staff: Mentor

No, logs are not needed.
Just solve for y in the first equation, then solve for y in the second equation, and equate the two expressions for y. If you know a and b, then you should be able to find c, and from that, you can find y.

Tips: Algebra equations typically use lower-case letters for variables. Also, don't use x for multiplication, especially in equations that already involve x (which your equations don't). Instead of writing AxY and BxY, just write AY and BY or ay and by.

3. Sep 28, 2017

### 1plus1is10

Okay... If B=A*Y and B=C/Y then C/Y=A*Y

Solution One:
A=(C/Y)/Y = C/(Y*Y) = C*1/(Y*Y) = 1/(Y*Y)*C
A/C=1/(Y*Y)
(A/C)/(Y*Y)=1
(A/C)*1/(Y*Y)=1
1/(Y*Y)=1/(A/C)
I'm clueless what happens next

Solution Two:
C=(A*Y)*Y = A*(Y*Y) = (Y*Y)*A
C/A=Y*Y
sqrt(C/A)=Y

Test:
sqrt(169/100)=1.3
Yeah!! I did it!!

Thanks Mark for pointing the way.

4. Sep 28, 2017

### 1plus1is10

5. Sep 28, 2017

### 1plus1is10

A/C=1/(Y*Y)
(A/C)*(Y*Y)=1
(Y*Y)*(A/C)=1
(Y*Y)=1/(A/C)
Y=sqrt(1/(A/C))
I did it again!!

6. Sep 29, 2017

### Staff: Mentor

I moved the thread to our homework section.

Note that -1.3 is a solution as well.

7. Sep 29, 2017

### 1plus1is10

mfb, this wasn't homework. LOL

Mark, my first instinct regarding pow was right:
I discovered that sqrt(x) is the same as pow(x, 1/2)
I mention this because last night while I was trying to sleep it came to me.
I thought... What if I need 2 or 3 or 4 numbers between the 2 I have (e.g. A,1,2,C or A,1,2,3,C or A,1,2,3,4,C etc.)?
Well, after a while, it clicked with me what the answer is: pow(x, 1/3) or pow(x, 1/4) or pow(x, 1/5) etc.

Thanks again though, because I wrote about "equate two equations" (a.k.a. "substitution method") in my personal Math Notes.
Very helpful - I never know when I will need it again.

8. Sep 29, 2017

### Staff: Mentor

But it's homework-like or could be a problem in a textbook, which means that the place to post it is here in the Homework & Coursework sections.
pow() is a standard library function in C and C++ and possibly a few other languages. If all you need is the square root, most languages have a library function called sqrt(). However, why are you bringing in programming functions to what is a fairly simple problem in algebra?

For this problem, neither would have been helpful. In solving for b, the next to last step is $b^2 = 169 * 100 = 16,900$. If you naively take the square root (or the 1/2 power), you get b = 130, but b = -130 is also a solution of the equation $b^2 = 16,900$. This means that $y = \pm 1.3$. That's what @mfb was talking about at the bottom of post #6.
I have no idea what you're trying to do here.
You aren't really "equating two equations" -- you're equating two expressions that have the same value. It doesn't make sense to set two equations equal to each other.

9. Sep 29, 2017

### 1plus1is10

I'm sorry you have gotten frustrated with me.

As for what I'm doing... Well to do my best at using math words: I'm "interpolating a power series" (sort of).
And yes, it's computer code - C++ in fact... I create charts - I'm visual.
Numbers, formulas, "Math", ugh! - I know it's right when I "see" it.
I wish I understood it more, but I don't - I accept that (I don't let it get me down).
I've got you, right?

(we're almost neighbors. I'm up here in Bellingham - hello there. My wife and I are heading down to Bellevue tomorrow to watch a movie - 11yr anniversary)

10. Oct 1, 2017

### Staff: Mentor

Hello, back. I'm a couple of counties south of you.