Solve for y?

Wej

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Can someone please help me solve this formula for y? I need it to be =y

((y*I11)/I19*D26-1)/((y*I11)/I19+1)=x

Thank you!
 
Kind of a mess.

Can you solve for Z in this expression?

(Z*D26-1)/(Z+1)=x

Note: I'm only guessing at your actual equation. Please add more parentheses, as necessary, to clarify intent.

For example, do you mean: [math]\dfrac{y\cdot I11}{I19} + 1[/math] or [math]\dfrac{y\cdot I11}{I19 + 1}[/math]?

The other piece (in the numerator) is even more potentially confusing.
 
I have no idea what you are even asking.

Is I11 supposed to be I11 or I * 11?

Is D26 supposed to be D26 or D * 26?

You have at least two pairs of unmatched parentheses.

When tkhunny said "kind of a mess," he was being very kind.
 
Sorry those are excel cell references. Let me try to clean that up.
 
Hopefully this is more clear. Thanks again for your help and I apologize for the confusion.

((y*A)/B*C-1)/((y*A)/B+1)=x

 
Hopefully this is more clear. Thanks again for your help and I apologize for the confusion.

((y*A)/B*C-1)/((y*A)/B+1)=x

Just to make sure, we need one more round of meaning checks. Does that mean this?

[MATH]\frac{\frac{y\times A}{B}\times C-1}{\frac{y\times A}{B}+1} = x[/MATH]​

(which is what it would mean in Excel), or

[MATH]\frac{\frac{y\times A}{B\times C-1}}{\frac{y\times A}{B+1}} = x[/MATH]​

or something else?
 
((y*A)/B*C-1)/((y*A)/B+1)=x Assuming it is \(\displaystyle \frac{\frac{yA}{BC}- 1}{\frac{yA}{B}+ 1}= x\) then the first thing I would do (because I really don't like fractions) is multiply on both sides by \(\displaystyle \frac{yA}{B}+ 1\):
\(\displaystyle \frac{yA}{BC}- 1= \frac{xyA}{B}+ x\)

To isolate y, Subtract \(\displaystyle \frac{xyA}{B}\) from both sides and add \(\displaystyle 1\) to both sides:
\(\displaystyle \frac{yA}{BC}- \frac{xyA}{B}= x+ 1\)
\(\displaystyle \frac{yA}{BC}- \frac{xyAC}{BC}= y\frac{A- AC}{BC}= x+ 1\)

I'll let you finish it.
 
1597418796914.png
This is the correct formula. I need to solve for Y. Sorry for the slow response.

Thank you!
 
Do as Halls suggested for the different problem: Multiply both sides by the denominator, then multiply both sides by the remaining LCD, B. Then show us some work so we can check it out.
 
I would substitute [math]\dfrac{yA}{B} \ with \ \ w.[/math]
So the problem becomes [math]\dfrac{wC - 1}{w+1}=x[/math]
I would then do as prof Halls suggested to solve for w. After solving for w you can substitute back and then easily solve for y.[/math]
 
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Is this correct?...
(A-AC/BC)/(X+1)=Y

Thanks again!
It is not what I get. I cannot see where we disgree because you did not show your work.

[MATH]\dfrac{\frac{ay}{b} * c - 1}{\frac{ay}{b} + 1} = x \implies \dfrac{\frac{1}{b} * (acy- b)}{\frac{1}{b} * (ay + b)} = x \implies[/MATH]
[MATH]\dfrac{acy - b}{ay + b} = x \implies acy - b = axy + bx \implies acy - axy = bx + b = b(x + 1) \implies [/MATH]
[MATH]y(ac - ax) = b(x + 1) \implies y = \dfrac{b(x + 1)}{ac - ax} = \dfrac{b(x + 1}{a(c - x)}.[/MATH]
Let's do a crude check.

If y = 2, a = 5, b = 10, and c = 7, then

[MATH]x = \dfrac{\frac{2 * 5}{10} * 7 - 1}{\frac{2*5}{10} + 1} = \dfrac{6}{2} = 3.[/MATH]
[MATH]\therefore y = \dfrac{10(3 + 1)}{5(7 - 3)} = \dfrac{10 * 4}{5 * 4} = 2 \ \checkmark.[/MATH]
 
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