Solving Fraction Issues: LCM vs Formula

[Amaz1ng]

Homework Statement

Simplify:
[tex]3c-(\frac{2a+c}{8}-\frac{a+2c}{4})[/tex]

The Attempt at a Solution

I got this:

[tex]\frac{9c}{8}[/tex]

But the answer is:

[tex]\frac{27c}{8}[/tex]

It's simply an issue with whether I find the LCM or I use this formula:

[tex]\frac{a}{b}-\frac{e}{f}=\frac{a*f-b*e}{f*b}[/tex]

If I use that, I get my answer which is wrong but if I find the LCM which is 8...it's the correct answer. What logic am I missing to know which to use that formula and when to just find the LCM?

 

[Mark44]

Homework Statement

Simplify:
[tex]3c-(\frac{2a+c}{8}-\frac{a-2c}{8})[/tex]

The Attempt at a Solution

I got this:

[tex]\frac{9c}{8}[/tex]

But the answer is:

[tex]\frac{27c}{8}[/tex]

I don't get either answer you posted. Are you sure you have written the problem exactly as it appears in your book?

It's simply an issue with whether I find the LCM or I use this formula:

[tex]\frac{a}{b}-\frac{e}{f}=\frac{a*f-b*e}{f*b}[/tex]

If I use that, I get my answer which is wrong but if I find the LCM which is 8...it's the correct answer. What logic am I missing to know which to use that formula and when to just find the LCM?

 

[Amaz1ng]

Opps...here it is.

Simplify:
[tex]3c-(\frac{2a+c}{8}-\frac{a+2c}{4})[/tex]

 

[SteamKing]

Are you sure you copies the expression in the OP correctly? The expression which appears actually simplifies to (21c - a)/8.

 

[Disconnected]

Are they both over 8 or is one of them over 16?

Nevermind! Just saw the new version.

Just multiply the last fraction by 2/2 to get
3c-(c-2a+c-2a-4c)/8

=3c-(-3c)/8

Then multiply the 3c by 8/8

=24c/8-(-3c)/8

=(24c+3c)/8

=27c/8

 

[Mark44]

I would get a common denominator for the two fractions and combine them. Then you could get a common denominator for your single fraction and the 3c term.

Where you are probably going astray is with the subtraction.
You can rewrite a difference such as this:
[tex]\frac{2a+c}{8}-\frac{a+2c}{4}[/tex]

as an addition, like so:
[tex]\frac{2a+c}{8} +\frac{-a-2c}{4}[/tex]

(Don't forget that you need to get the common denominator.)

 

[Disconnected]

I really need to get used to using latex on this site.

 

[VietDao29]

Opps...here it is.

Simplify:
[tex]3c-(\frac{2a+c}{8}-\frac{a+2c}{4})[/tex]

a - (b - c) = a - b + c.

Or in general, if there's a plus sign in front of a pair of parentheses, then you don't have to change the sign of each elements in those prentheses.

But if it's a minus sign, then you have to change the sign of every element in those parentheses. Like this:

a + (b - c + d - e) = a + b - c + d - e.

a - (b - c + d - e) = a - b + c - d + e.

Now, do the calculations again. And see if you get the later result. :)

 

[Amaz1ng]

I don't think you guys understand what I'm saying. I'm saying that when I use the LCM method it's different from when I use the cross multiplication method.

But shouldn't it yield the same result? I'm obviously missing the logic of the cross multiplication method.

...which is this:
[tex]
\frac{a}{b}-\frac{e}{f}=\frac{a*f-b*e}{f*b}
[/tex]

 

[Mark44]

I don't think you guys understand what I'm saying. I'm saying that when I use the LCM method it's different from when I use the cross multiplication method.

But shouldn't it yield the same result? I'm obviously missing the logic of the cross multiplication method.

...which is this:
[tex]
\frac{a}{b}-\frac{e}{f}=\frac{a*f-b*e}{f*b}
[/tex]

Cross multiplication doesn't apply here, since you are not trying to solve an equation, but are instead trying to simplify a rational expression.

This is cross multiplication:
[tex]
\frac{a}{b} = \frac{c}{d}
[/tex]
[tex]
\Rightarrow ad = bc
[/tex]
This is equivalent to multiplying both sides of the equation by bd and simplifying the resulting fractions.

 

[Amaz1ng]

Ok yeah but there's another method that looks like this right:

[tex]
\frac{a}{b}-\frac{e}{f}=\frac{a*f-b*e}{f*b}
[/tex]

So I'm trying to find out when that's applicable and when it's not.

 

[Mark44]

Here you're just finding the least common denominator. In the first fraction, you're multiplying by 1 in the form of f/f. In the second, you're multiplying by 1 in the form of b/b. You can always multiply by 1.

The fraction on the right is the same as (af)/(bf) - (be)/(bf).

 

[Dead Boss]

I tried it both ways and I got the correct result 27c/8. You must have made some mistake.

 

[mege]

Don't forget the 3c hanging out in the beginning needs to have the same denominator as well for your problem.

Cross Multiplication

Cross multiplication is useful for an EQUATION involving two fractions set equal to each other (an equation remember is something with an equal's sign in it to start). Generally with equations at this level of math, you're solving for a piece of the puzzle and 'filling in' a variable in the equation.

eg:

[tex]\frac{z}{8} = \frac{3}{4}[/tex]

What does z come out to? This is a good cross multiplying candidate. Cross multiplying we get

[tex]
z*4 = 8*3
[/tex]
[tex]
z*4 = 24
[/tex]
(divide each side by 4)
[tex]
z = 6
[/tex]

Even if one side doesn't have a fraction, the idea of cross multiplying is an easy way to remember it

[tex]
\frac{k}{2} = 12
[/tex]
[tex]
\frac{k}{2} = \frac{12}{1}
[/tex]
[tex]
k*1 = 2*12
[/tex]
[tex]
k=24
[/tex]

(you could also just multiply both sides by 2 from the start - which has the exact same result, for some reason, even late in college, I think about equations like this in terms of cross multiplying even so)

Fraction Addition/Subtraction

The function that you reference below is good for evaluating an expression (an expression doesn't need an equals to start, since you're always finding the result). Below is just a shortcut to find what you need when dealing with fraction subtraction or addition.

[tex]\frac{a}{b}-\frac{e}{f}=\frac{a*f-b*e}{f*b}[/tex]

Probably the first thing you learned with fraction math is that addition and subtraction requires the same denominator.

[tex]\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}[/tex]

Whenever you see two fractions of the same denominator - you can easilly apply the above to addition or subtraction. Now, how do we solve for mixed denominators. We can always multiply any number (or expression) by 1 and its value stays the same. Here's some ways to rewrite 1:

[tex]1 = \frac{a}{a} = \frac{ab}{ab} = \frac{6b}{6b} = \frac{b}{b} = \frac{4}{4} = \frac{1}{1}[/tex]

Now, we'll use that multiplication of 1 a few times to set our denominators equal.

[tex]
\frac{a}{b}-\frac{e}{f}=?
[/tex]
[tex]
\frac{a}{b}*1-\frac{e}{f}*1
[/tex]
[tex]
\frac{a}{b}*\frac{f}{f}-\frac{e}{f}*\frac{b}{b}
[/tex]
We're just multiplying each of the fractions by 1, and have NOT changed the value of each expression. We decided what variables to use based on what other fraction we needed to compare it to. With fraction multiplication we can just multiply above and below the bar to get:

[tex]
\frac{a*f}{b*f}-\frac{e*b}{f*b} = \frac{a*f-b*e}{f*b}
[/tex]

Does this long explanation of it help a little more?