Solve Math Question: Rearrange Formula to Make y Subject

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The discussion revolves around rearranging the equation ax + by = cx - dy to isolate y. Participants emphasize the importance of using properties of equality, specifically additive and multiplicative inverses, to manipulate the equation. The steps include adding dy and subtracting ax to isolate y, leading to the expression (b + d)y = (c - a)x. Finally, dividing by (b + d) results in y = ((c - a)/(b + d))x, with the condition that b + d ≠ 0. The conversation also touches on the need for clarity in mathematical terminology and the importance of following forum guidelines for homework help.
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I am having trouble answering a question given to me in maths:

Rearrange this formula to make y the subject.

ax + by = cx - dy

Any help would be greatly appreciated, plus the steps taken to get the answer.
 
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What do you exactly mean, "y the subject"? Are you interested in English grammar, or are you interested in introductory Algebra? you seem to be asking for a formula for y. So your general interest is to practice properties of equality.

Use additive inverse of -dy;
use additive inverse of ax;
use distributive property;
use multiplicative inverse property.
 
Symbolipoint, "make y the subject of the equation" is "English" English for what us "Ammuricans" would call "solve the equation for y".

Willy Vee, you need to get y by itself on one side of the equation: y= ...

To do that you look at what has been "done to y" and then do the opposite. That is why symbolipoint has three "inverses"!

You have, to start with, ax+ by= cx- dy. You want y only on the left side of the equation and right now there is a "-dy" on the right. Do the opposite- add dy (dy is the "additive inverse of -dy) to both sides: ax+ by+ dy= cx- dy+ dy= cx. You want no x on the left so add -ax to both sides (-ax is the "additive inverse of ax): ax+ by+ dy- ax= by+ dy= cx- ax. You should see that you can now factor y out of the left side: by+ dy= (a+d)y so you really have (a+d)y= cx-ax. If a+ d is not 0, then we can divide both sides by a+d (or multiply by the "multiplicative inverse of a+d, 1/(a+d)) to get
y= (cx- ax)/(a+ d). You might also notice that you can factor an x out of that: [(c-a)/(a+d)]x.
 
You left the b out HallsofIvy ;)

First add dy and subtract ax from each side to isolate y on the left.

by+dy=cx-ax

Take a common factor of y from the right side and an x from the left side.

(b+d)y=(c-a)x

Divide each side by (b+d), and you have your answer

y=( (c-a)/(b+d) )x and b+d≠0
 
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symbolipoint said:
What do you exactly mean, "y the subject"? Are you interested in English grammar, or are you interested in introductory Algebra? you seem to be asking for a formula for y. So your general interest is to practice properties of equality.

Use additive inverse of -dy;
use additive inverse of ax;
use distributive property;
use multiplicative inverse property.

cor you sound like a proper bore!...not to be mean or nothing
 
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nah I'm stuck on some rearrange formula homeowrk - teacher expects me to do it even though i weren't there
 
fifim007 said:
cor you sound like a proper bore!



...not to be mean or nothing
You just can't help it?

Ok, so that I don't bore you anymore I will not respond to any more of your questions.

In fact, to be helpful, I will warn others not to give detailed answers to your questions since they bore you.

And, since it was clearly to0 boring for you to read the info you were required to when you registered (in particular the part that says homework must go in the homework section and you must show an effort), I will tell you that here (hope fifim007 got this far before she got bored) and move this to the PreCalculus homework site.
 
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