Expert: Josh - 8/30/2009

Question
QUESTION: Hi there,

first off thank you for your time.

My question is regarding solving the following equation:

(x-5)/4  +  (3)/2  =  (x+2)/3

I have no idea how to get rid of the denominator.

Some say you have to find the LCD, which is 12, but then they explain nothing.

Do you multiply (x-5)/4   bot the numerator and den. by 3?  so you get  3(x-5)/12  ?

How do you then get rid of the denominator?

Some say, like the textbook, just multiply everything by 12. But I don't understand this at all.

In short, I understand almost everything, but how you get rid of the denominator.   

Can you please explain, step by step in detail what you multiply by what, how you got that, how it makes the denominator dissappear, etc?

Thank you very much, and I really hope you can help me in this. I am so frustrated, nobody seems to be able to explain this well enough.

ANSWER: Hi Stephen,

I think I know the missing link. To understand this properly, we will first look at the problem, then revisit an important observation which enables us to solve the problem.

ANALYZE THE PROBLEM: The difficulty we have is that fractions do not add directly when they have different denominators. i.e., we cannot just add the top expression in each fraction together to obtain the numerator, then retain or attempt to combine the denominators. This works only when the fractions have the SAME denominator.

MOTIVATION: An important observation is that a fraction (say, A/B) always maintains the same ratio when we multiply it by 1. Now, the identity "1" can be written in a variety of ways. Regardless of the value of X, X/X is always 1. So, we can multiply (A/B) by (X/X), and (A*X)/(B*X) still represents the SAME proportion as (A/B). As an example, let A/B be 2/7. If we pick X to be 3, then, (2/7)*1 = (2/7)*(3/3) = (2*3)/(7*3) = 6/21 represents the same proportion as the original fraction 2/7.

Why is this useful to know?
EXAMPLE 1: Suppose we want to add (A/B)=2/7 to (C/D)=1/21. To simplify 2/7 + 1/21, we once again run into the problem that the two fractions cannot be added directly because the denominators are different. We can overcome this difficulty if we can manage to find a multiplier "X", such that when (A/B) is multiplied by X/X, its denominator (B*X) is same as D [the denominator of C/D]. In other words, we want 7*X to be equal to 21. X=3 would do. So, we are allowed to write (A/B) as (A/B)(X/X) without changing its proportion. We say that 2/7 is equivalent to (2/7)*(3/3)=(2*3)/(7*3)=6/21. Now, replacing 2/7 with 6/21, we are in a position to add it directly with 1/21. In summary,
2/7 + 1/21
= (2/7)*(3/3) + 1/21
= (2*3)/(7*3) + 1/21
= 6/21 + 1/21
= 7/21    dividing top by 7 and bottom by 7 gives
= 1/3

EXAMPLE 2: Consider 2/3 + 1/4. In this case, the denominator of the second fraction "4" cannot be divided by the denominator of the first fraction "3". So, we have to multiply BOTH fractions by some identity, to ensure that both fractions in (2/3)*(X/X) + (1/4)*(Y/Y) have the SAME denominator. Generally, we go for the smallest values (in this case, X and Y) that we can find to make the denominators compatible. A simple choice here would be X=4 and Y=3. Essentially, we multiply the top and bottom expressions in the first fraction, by the denominator of the second fraction, and vice versa (we multiply the top and bottom bits in the second fraction, by the denominator of the first fraction). In summary,
2/3 + 1/4
= (2/3)*(4/4) + (1/4)*(3/3)
= (2*4)/(3*4) + (1*3)/(4*3) note: both fractions now have a common denominator
= 8/12 + 3/12
= (8+3)/12
= 11/12

RETURNING TO YOUR QUESTION: (x-5)/4 + 3/2 = (x+2)/3

1) Problem: Each fraction has a different denominator. Thus, we cannot add them directly together.
2) Objective: Rewrite each fraction without changing its proportion. Multiply each fraction by an identity such that each fraction shares a common denominator. This is basically an extension of Example 2.
3) Specifically, find W,Y and Z such that (x-5)*W/(4*W) + (3*Y)/(2*Y) = (x+2)*Z/(3*Z) all share the same denominator (a least common denomintor, or LCD, if you prefer). We focus only on the denominators at this stage. We want 4W = 2Y = 3Z.

With practice, you will be able to do this by inspection. Right now, we will analyze this the long way.
Let's suppose M is the LCD. By definition, the Least Common Denominator is the smallest integer that is divisible by the denominator of all given fractions. So, our task is equivalent to asking: (i) what do I have to multiply (the first denominator) 4 by to get M? (ii) what do I have to multiply (the second denominator) 2 by to get M? (iii) what do I have to multiply (the third denominator) 3 by to get M.

Looking at the denominators {4, 2, 3}, from the perspective of the first denominator (the number "4"), part (i) may be rephrased as "what prime factors present amongst the other denominators am I currently missing". Well, from 4=2*2, it is definitely divisible by 2 (the second denominator in the list). But, the third denominator "3" is clearly missing. So, we let W=3. This gives a least common multiple of M=4*W=12.

From the perspective of the second denominator (the number "2"), we again look for the missing prime factors present amongst the other denominators that are currently missing. Comparing itself (2) with the first denominator 4=2*2, it is missing a single "2". Comparing itself (2) with the third denominator "3", it is missing 3. So, we need Y=2*3 (the product of both missing prime factors) so that 2*Y=12 equals the LCM.

Finally, from the perspective of the third denominator (the number "3"), we again look for the missing prime factors present amongst the other denominators that are currently missing. Comparing itself (3) with the first denominator 4=2*2, it is missing a "2*2". Comparing itself (3) with the second denominator "2", it no longer needs a two since it will have two 2's. So, we set Z=2*2 (the product of missing prime factors) so that 3*Z=12 equals the LCM.

With experience, you may think that this elaborate procedure is an over-kill. But this general strategy always works. All this amounts to is saying:
(x-5)/4 = (x-5)/(4) * (W/W) where W=3
       = (x-5)*3/(4*3)
       = 3(x-5)/12
| Comment: so you had the right idea when you asked "Do you multiply (x-5)/4
| both the numerator and den. by 3? ..."

3/2 = (3/2) * (Y/Y) where Y=2*3=6
   = (3*6)/(2*6)
   = 18/12  (note: Y is chosen so that this fraction now has a common denominator of 12)

(x+2)/3 = (x+2)/3 * (Z/Z) where Z=2*2=4
       = (x+2)*4/(3*4)
       = 4(x+2)/12

So, the original sum of fractions is reduced as follows:
(x-5)/4 + 3/2 + (x+2)/3
= 3(x-5)/12 +(18/12) + 4(x+2)/12
= [3(x-5) + 18 + 4(x+2)] / 12
= [3x-15 +18 +4x+8]/12
= (7x+11)/12 ...(*)

REMARK: There is generally no harm done using a common denominator other than the smallest one (unless you are specifically asked to do so in the question). If it just asks you to simplify the expression, you may if you wish, multiply each fraction by an identity, say (W/W), where W comes from the product of denominators from fractions other than its own. This may require simplification later, but everything is legitimate. As an example, for (x-5)/4 + 3/2 + (x+2)/3, you can multiply {4,2,3} together to get a common multiple of M=4*2*3=24. (This is not the lowest common multiple, as we have shown before, the LCD is 12 for this question). Our next concern is what do we multiply each fraction with?

For the first denominator 4, 24/4=6. So, we can multiply (x-5)/4 with (W/W) where W=6.
For the first denominator 2, 24/2=12. So, we can multiply 3/2 with (Y/Y) where Y=12.
For the first denominator 3, 24/3=8. So, we can multiply (x+2)/3 with (Z/Z) where Z=8.

Then, (x-5)/4 + 3/2 + (x+2)/3
= 6(x-5)/24 +(36/24) + 8(x+2)/24
= [6(x-5) + 36 + 8(x+2)] / 24
= [6x-30 +36 +8x+16]/24
= (14x+22)/24    ...which may be factorized as
= 2(7x+11)/(2*12)...compare this to the line (*) before, we cancel out the common factors to get
= (7x+11)/12

Re: "how you get rid of the denominator"
The objective is not so much as getting rid of the denominator, but expressing each fraction in such a way that they share a common denominator. The rationale is simple, if I ordered two pizza (both identical in size and flavor) and I cut the first into three equal portions and I cut the second into nine equal portions, one slide from the first pizza is certainly not compatible with a slide from the second pizza. The former represents 1/3, while the latter represents 1/9. If we ignore the relative proportions, and add them blindly it makes no sense. However, if we interpret 1/3 as equivalent to 3/9, then they are on equal footing. We can then say that 1/3=(3/9) is three times larger than 1/9 (the smaller slice from the second pizza). The denominator tells us what the "1" is relative to.

This has been rather long. But I hope it helps you clear some of the confusion.

---------- FOLLOW-UP ----------

QUESTION: Great answer! Thank you.

My only other question is, and this is where I struggle the most, is how to eliminate the denominator.

The solution my professor gave me is:

3(x-5) + 18 = 4(x+2)
x= -5

How did she eliminate the denominator right off the bat, and how would I go about finishing the problem the way you started it?

thank you

Answer
Sorry Stephen,

I misread the question. Here, instead of adding three fractions, we have an equation where each of the fractions has a different denominator.

Now that we know how to find a common denominator (using the prime factorization method), we should be able to go from
(x-5)/4 + 3/2 = (x+2)/3 to 3(x-5)/12 +(18/12) = 4(x+2)/12

i.e., [3(x-5) + 18]/12 = 4(x+2)/ 12
At this point, because the denominators are the same on both sides of the equation, like in the pizza example, the proportions can be directly compared to one another since they are on equal footing. Whenever we have an equation, and the denominator on both sides of the equation are the same, we can just ignore the denominator altogether. Equality holds provided the numerator expression on the LHS equals to the numerator on the RHS of the equation. If you are not yet comfortable with this, you can multiply both sides by 12 to formally cancel out the "12" in the denominator. This is one of the rules of algebra. We can multiply both sides of an equation by a constant, without changing the logic of the equation. Carrying on,

[3(x-5) + 18]*12/12 = 4(x+2)*12/12
3(x-5) + 18 = 4(x+2)
3x-15 + 18 = 4x+8
-15+18-8 = 4x-3x
x=-5

Let me know if anything is unclear.