Solve x/a + y/b + z/c = 1 for A

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
burnbird16
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Solve x/a + y/b + z/c = 1 for A

x/a + y/b + z/c = 1.

Why oh WHY did my calculus teacher give me a summer assignment! *tears hair out* Someone help with this problem? I also have another that's a lot like it if you can help with that one also, still solving for A.

V = 2(ab + bc + ca)

stapel_eliz
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burnbird16 wrote:Why oh WHY did my calculus teacher give me a summer assignment!

To give you something with which to review over the summer, in hopes of preventing the loss of two to three months' worth of learning, as usually happens.

burnbird16 wrote:x/a + y/b + z/c = 1.

How does the stipulated variable $A$ relate to the variables $a$, $b$, $c$, $x$, $y$, and $z?$

burnbird16 wrote:still solving for A.

V = 2(ab + bc + ca)

Same question.

When you reply, please include a clear listing of your work so far, or stipulate that you first need lesson instruction before you can get started. Thank you!

burnbird16
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Re: Solve x/a + y/b + z/c = 1 for A

Er...I didn't need a rant, just some help with a math problem or two. o.o

I suppose I'll need a lesson in both of these, as I really don't understand what you mean by "stipulated variable" and how it "relates", but they seem really familiar from Pre-Cal, but I can't think of a single shortcut to do them.

Here's what I got for the first one:

x/a + y/b + z/c = 1 <-- I started by condensing the left side into one big fraction.

(bcx + acy + abz)/abc = 1 <-- Next, I got rid of the denominator "abc" and moved it over to the other side, using multiplication.

bcx + acy +abz = abc <-- Then, I took "acy" and "abz" both of which contain the variable "a" which I'm solving for, and subtracted them from both sides.

bcx = abc - acy - abz <-- Then, I pulled out a, in reverse distribution.

bcx = a (bc - cy - bz) <-- Finally, I divided both sides by the large trinomial, leaving a alone on one side, so my answer was

bcx/(bc - cy - bz) = a

When that one was done, I did the next one, and got an answer:

V = 2 (ab + bc + ca) <-- Okay, volume problem, and I think the last one was law of sines, if I remember correctly. I began by distributing the 2

V = 2ab + 2bc + 2ca <-- Next, like the last problem, I moved anything that lacked the "a" variable over to the other side with "V".

V - 2bc = 2ab + 2ca <-- Then I pulled out variable "a".

V - 2bc = a (2b + 2c) <-- Finally, I divided the binomial on the right from both sides, leaving me with an answer.

(V - 2bc)/(2b + 2c) = a

burnbird16
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Re:

How does the stipulated variable $A$ relate to the variables $a$

Sorry, perhaps I wasn't clear. $A$ and $a$ are the same variable, I accidentally capitalized one and not the other. That was my mistake.

stapel_eliz
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burnbird16 wrote:Er...I didn't need a rant...

I'm sorry to hear that somebody "ranted" at you. (That must have been a private message, since all that's been done publically is request clarification, so any suggestions and helps will be correct. Please contact the Admin about whatever message you received. Thanks!)

burnbird16 wrote:I suppose I'll need a lesson in both of these....

To learn how to solve literal equations (that is, equations with more than one variable, and you're solving for a stipulated variable, such as $A$, in this case), try this lesson. (But from the work you've shown, I don't think you are having any difficulty with the topic. It's a possibly poorly-written assignment which is the problem.)

Since they've given you no relationship between $A$ and any of $a$, $b$, $c$, $x$, $y$, and $z$, I don't see how you can proceed to the answer they seem to want.

burnbird16 wrote:Here's what I got for the first one....

bcx = abc - acy - abz <-- Then, I pulled out a, in reverse distribution. . .<== Excellent!

Excellent! That's the step that most students miss!

bcx/(bc - cy - bz) = a

That solves the literal equation for $a$, but you need to solve it for $A$. Unless your instructor typoed the instructions, what you've done, though completely correct and very-well documented, cannot be the expected answer. Same thing for the second exercise: your work is great, and I get the same answer for "$a\, =$", but your instructor asked you for "$A\, =$". I have no idea what you're supposed to do next.

burnbird16
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Re: Solve x/a + y/b + z/c = 1 for A

Thanks for your help there, though, as I said, there was no problem with the assignment, merely my own typo - I typed A where I ought to have put "a".

But here's another one in the same area I'm having trouble with. Here, we're solving for positive r. The equation is that of the surface area of a sphere.

A = 2 pi r^2 + 2 pi rh, solve for positive r.

I got to the point where I pulled out 2pi, divided it from both sides, and had

A/2pi = r^2 + rh

...but I don't know where to go from there. I'm completely lost. What do I do?

stapel_eliz
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burnbird16 wrote:...as I said, there was no problem with the assignment, merely my own typo

Must have been a server or browser error (or I'd left a window open way longer than I'd realized; I could only see the first reply. Sorry!

burnbird16 wrote:A = 2 pi r^2 + 2 pi rh, solve for positive r.

So you've got:

. . . . .$A\, =\, 2 \pi r^2\, +\, 2 \pi rh$

...and you need to solve this for $r\, >\, 0$. So first let's rearrange:

. . . . .$0\, =\, 2 \pi r^2\, +\, 2 \pi rh\, -\, A$

. . . . .$0\, =\, \left(2 \pi\right)r^2\, +\, \left(2 \pi h\right)r\, -\, A$

So this is a quadratic in $r$, and you can solve this using the Quadratic Formula, with $a\, =\, 2 \pi$, $b\, =\, 2 \pi h$, and $c\, =\, -A$. Obviously, only the "plus" part of the square root can give you a positive answer.

burnbird16
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Re: Solve x/a + y/b + z/c = 1 for A

You are a GENIUS! Thank you so much, that makes perfect sense. I'll be sure to come back if I encounter any more odd questions. Thanks!