Helping Daughter w/ math, this one is stumping me: 2=(ab+c)/a Solve for a.

Narcosys

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Usually pretty good at math, but I keep getting stumped and getting a variable equals a variable.

Problem:

2=(ab+c)/a

Solve for a.

Been trying to solve for b and c cause I can isolate those, can't seem to isolate a.

Thanks for any help.
 

Subhotosh Khan

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Usually pretty good at math, but I keep getting stumped and getting a variable equals a variable.

Problem:

2=(ab+c)/a

Solve for a.

Been trying to solve for b and c cause I can isolate those, can't seem to isolate a.

Thanks for any help.
Can you please post the problem "exactly" as it was presented to your daughter.
 

Denis

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Dr.Peterson

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Usually pretty good at math, but I keep getting stumped and getting a variable equals a variable.

Problem:

2=(ab+c)/a

Solve for a.

Been trying to solve for b and c cause I can isolate those, can't seem to isolate a.

Thanks for any help.
Please show your work, so we can see where you are having trouble.

I would first multiply both sides by a, and then collect terms containing a on one side. Many students get stuck at that point; the key word is "factor".
 

Narcosys

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That is exactly as it is presented.

I get you start out multiplying by a to get
2a=ab+c
Which is also equivalent to 2= b + (c/a)

However let's go with the former
2a = ab+c

I can divide by a to get the same as the latter 2= b + (c/a)
Then solve for b to get b = c/a - 2

2 = (a(c/a - 2)+ c)/a
Multiply by a and then distribute the a that was in the numerator

2a= c/a^2 - 2a + c
Negate the 2a on the right

4a = c/a^2 + c

And that's different than what I've gotten in the past.

Or should I have started out cancelling the a that's divided a
2 = (a(c/a - 2) + c)/a

2= c/a -2 + c/a

2= 2(c/a) - 2

4= 2(c/a)

2= c/a

2a = c

Plug in both results for b & c

2= (a(2a/a - 2) + 2a)/ a
The dividing a negates the distributing a

2=2a/a - 2 + 2a/a
4 = 2a/a + 2a/a
4 = 2a
2 = a

????
 

Narcosys

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tkhunny

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That is exactly as it is presented.

I get you start out multiplying by a to get
2a=ab+c
Which is also equivalent to 2= b + (c/a)
Important hint: After multiplying by 'a', please don't even consider next dividing by 'a'. That makes absolutely NO sense.

You are trying to solve for "a". Normally, avoid things that make it worse.

You have 2a = ab + c

Your task is to solve of "a". Typically, this requires collecting multiple appearances of "a". Keep the end in mind.
 

Jomo

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Usually pretty good at math, but I keep getting stumped and getting a variable equals a variable.

Problem:

2=(ab+c)/a

Solve for a.

Been trying to solve for b and c cause I can isolate those, can't seem to isolate a.

Thanks for any help.
Solving for b and/or c will not help. How would you solve for a if b=3 and c=4. Then do the same steps using b and c.
 

Narcosys

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Solving for b and/or c will not help. How would you solve for a if b=3 and c=4. Then do the same steps using b and c.

Solving for b leaves only a and c variables on one side. You then throw those in for b, leaving only a and c variables in the equation and you can solve for c, leaving only a variables. Tjen you attempt to solve for a.
 

Narcosys

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Important hint: After multiplying by 'a', please don't even consider next dividing by 'a'. That makes absolutely NO sense.

You are trying to solve for "a". Normally, avoid things that make it worse.

You have 2a = ab + c

Your task is to solve of "a". Typically, this requires collecting multiple appearances of "a". Keep the end in mind.
The dividing by a would negate the ab

But I get what you're saying and forgot about the factoring of distributive properties

2a= ab + c

2a - ab = c

a(2 - b) = c

a = c /(2 - b)


Thanks.
 

Dr.Peterson

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Solving for b leaves only a and c variables on one side. You then throw those in for b, leaving only a and c variables in the equation and you can solve for c, leaving only a variables. Tjen you attempt to solve for a.
It sounds as if you are confusing this with a system of equations, where you can use one equation to eliminate a variable in another. Here you have only one equation, and the answer will be an expression containing b and c.

Have you tried doing what was suggested? You have 2a = ab + c, which is better than the original because a is not in the denominator. Now you have to gather terms with a on one side, and then, as I hinted, factor a out of two terms, so that there is only one a in the equation.
 

Denis

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JeffM

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Usually pretty good at math, but I keep getting stumped and getting a variable equals a variable.

Problem:

2=(ab+c)/a

Solve for a.

Been trying to solve for b and c cause I can isolate those, can't seem to isolate a.

Thanks for any help.
"Solve for a" does not necessarily mean finding a numeric value for a. It means finding an expression that equals a. The simplest type of expression is indeed a numeral, but it may not be possible to get the simplest type.

A general rule is that you need at least n equations to get numeric answers if you have n variables. You have 3 variables and only one equation. So the best that you can do is to find the simplest algebraic expression that equals the indicated variable.

Solve for p given \(\displaystyle r = 6q - \dfrac{3}{2p}.\)

Three variables, one equation means an algebraic solution, not a numeric one.

\(\displaystyle r = 6q - \dfrac{3}{2p} \implies \)

\(\displaystyle 2p * r = 2p * \left ( 6q - \dfrac{3}{2p} \right ) \implies \)

\(\displaystyle 2pr = 12pq - 3 \implies \)

\(\displaystyle 12pq - 3 = 2pr \implies \)

\(\displaystyle 12pq - 3 + 3 = 2pr + 3 \implies \)

\(\displaystyle 12pq = 3 + 2pr \implies \)

\(\displaystyle 12pq - 2pr = 3 + 2pr - 2pr \implies\)

\(\displaystyle 12pq - 2pr = 3\)

\(\displaystyle p(12q - 2r) = 3 \implies \)

\(\displaystyle \dfrac{p(12q - 2r)}{12q - 2r} = \dfrac{3}{12q - 2r } \implies\)

\(\displaystyle p = \dfrac{3}{12q - 2r} = \dfrac{3}{2(6q-r)}.\)

Now check your work.

\(\displaystyle 6q - \dfrac{3}{2p} = 6q - \dfrac{3}{2 * \dfrac{3}{2(6q - r)}} = \)

\(\displaystyle 6q - \dfrac{\dfrac{3}{1}}{\dfrac{3}{6q - r}} = 6q - \dfrac{3}{1} * \dfrac{6q - r}{3} = \)

\(\displaystyle 6q - (6q - r) = 6q - 6q + r = r.\)

It checks.

Now you have a model.
 
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