Question: Equation with Many Fractions

Comment on Equation with Many Fractions

The last example is wrong, as far as I have solved it. The result should be -8/5 and not -5/8...
greenlight-admin's picture

Can you show your calculations?
Once we get to -5 = 8x, we divide both sides by 8 to get: -5/8 = 8x/8
Simplify to get: -5/8 = x

Why can you multiply both sides by 20x if one of the denominators (5 in the second fraction on the left) does not include x?
greenlight-admin's picture

Yes, it's okay to do that.
When we do this, the only thing we need to be sure of is that x does not equal zero.
We can be certain that x does not equal zero, since the original equation wouldn't make any sense if x equaled zero.

Why did you choose to multiply both sides by 20x instead of 20? I understand that you can, but why did you choose to include the "x"?
greenlight-admin's picture

Hi Garrett,

In order to eliminate all fractions, we must multiply both sides of the equation by the least common multiple of all of the denominators (2x, 5 and 4x).
For more on this see:

So, 20x is the least common multiple of 2x, 5 and 4x

Notice what happens if we just multiply both sides by 20.

We get: 20(1/2x + 3/5) = 20(1 + 3/4x)
Expand: 20/2x + 60/5 = 20 + 60/4x
Simplify: 10/x + 12 = 20 + 15/x

As you can see, we still have some fractions because we were unable to cancel out the x's in the denominators.

If we want to eliminate the existing fractions, we must multiply both sides by x.
We get: x(10/x + 12) = x(20 + 15/x)
Expand: 10x/x + 12x = 20x + 15x/x
Simplify: 10 + 12x = 20x + 15

So, rather than multiply both sides of the equation first by 20 and then later by x, we could have saved a step by multiplying both sides by 20x.

Does that help?


I don't understand how you got -5/8. I looked back at how you carried the x and I think it was supposed to be 10x+12=20+15x instead of 10+12x=20x+15. Is this correct? If so, wouldn't the answer change to -8/5 which would mean that B would be the correct answer? Sorry if this is a dumb question.
greenlight-admin's picture

Let's take it step-by-step. When it comes time to multiple both sides by 20x (aka 20x/1), we'll show an extra step.

Given: 1/2x + 3/5 = 1 + 3/4x
Multiple both sides by 20x/1 to get: 20x/2x + 60x/5 = 20x + 60x/4x
Simplify: 10 + 12x = 20x + 15
Subtract 10 from both sides: 12x = 20x + 5
Subtract 20X from both sides: -8x = 5
Divide both sides by -8: x = -5/8

Does that help?

Yes, thanks!

my confusion is dropping the x from 20x/2x. is it not 10x?
greenlight-admin's picture

Let's use a nice fraction property that says ab/cd = (a/c)(b/d)
So, we can write: 20x/2x = (20/2)(x/x)
= (10)(1)
= 10

Aside: We also used the fact that x/x = 1

and the 60x/4x. why not 15x. Thanks
greenlight-admin's picture

We'll use the same fraction property: ab/cd = (a/c)(b/d)
So, we can write: 60x/4x = (60/4)(x/x)
= (15)(1)
= 15

Another way to verify whether 60x/4x = 15 is to TEST some values of x.

Try x = 2
60x/4x = 60(2)/4(2) = 120/8 = 15 PERFECT!

Try x = 3
60x/4x = 60(3)/4(3) = 180/12 = 15 PERFECT!

Try x = 10
60x/4x = 60(10)/4(10) = 600/40 = 15 PERFECT!

Since you are suggesting that 60x/4x = 15x, let's test this out by testing some more values

Try x = 2
Does 60x/4x = 15x?
Plug in x = 2 to get: 60(2)/4(2) = 15(2)
Evaluate both sides to get: 15 = 30
Doesn't work. So, it is not the case that 60x/4x = 15x

Does that help?

That helps. Thanks

When I added 1/2x + 3/5 I got 6x+5/10x. And 1+ (3/4x) is (4x+3)/4x. Setting these equal to each other and cross multiplying, I got 4x*(6x+5) = 10x*(4x+3). After simplifying I get the quadratic equation 16(x^2) +10x+0=0. This has two solutions, 0 and -5/8. Why do I get two solutions instead of only one like you did with your method? Thanks.
greenlight-admin's picture

Great question! First off, your solution strategy is perfectly valid (although perhaps longer).

We get x = 0 as a solution due to the fact that, if x = 0, the fractions on each side are UNDEFINED (since we're dividing by 0). The equation you created "behaves" as though x = 0 is a solution, but we must recognize that it's actually an invalid solution.

I went at this in a completely different way using equivalent fractions. 1/2x became 2/4x and 1 became 5/5. I was then able to just add and subtract to get -1/4x = 2/5. I then cross multiplied to get -5=8x then divided by 8 to get x = -5/8.
greenlight-admin's picture


Hi, I did my math the same as Dawnhughan using equivalent fractions, however instead of cross multiplying at the last step, I multiplied both sides by -4. This gave me and answer of x= -8/5 instead of the correct answer, -5/8. Why is this wrong?
greenlight-admin's picture

In the future, it would help if you showed all of your calculations. Otherwise, I'm forced to guess the steps you took to get x = -8/5.

Here's my best guess at what you did.

Take: 1/2x + 3/5 = 1 + 3/4x
Rewrite first term as: 2/4x + 3/5 = 1 + 3/4x
Subtract 3/4x from both sides: -1/4x + 3/5 = 1
Subtract 3/5 from both sides: -1/4x = -2/5
Multiply from both sides by -4 to get: 1/x = -8/5

At this point, we know that 1/x = -8/5, however we want to find the value of x (not 1/x)
Take: 1/x = -8/5
Invert both sides to get: x/1 = -5/8
In other words, x = -5/8

Is that what happened in your solution?


Hello, you said in a previous video that for matching operations we cannot multiply or divide by variables unless we are 100% sure that the variable is positive. How can we then now multiply by 20x? Thanks :D
greenlight-admin's picture

Hi Julian,

That's not quite what I said.

In Quantitative Comparison questions, you cannot multiply/divide the TWO QUANTITIES (Quantity A and Quantity B) by variables unless we are 100% sure that the variable is positive..

In this question, we are multiplying the given EQUATION by 20x. This is okay.

Here's the video that says you cannot multiply/divide the TWO QUANTITIES by variables unless we're sure the variable is positive:

Here's the video on solving EQUATIONS through matching operations:

Does that help?


Thank you this reply was my eureka moment !!!
greenlight-admin's picture

Awesome - thanks for the feedback!

What's the least common multiple of 2x and 5 only? And also why 20x why not just 20 since 5 doesn't have an X
greenlight-admin's picture

The least common multiple of 2x and 5 is 10x. Here's why:

Our goal here is to find a value that will help cancel out all of the numerators (i.e., eliminate all of the pesky fractions).

One way to think of it is in terms of find the lowest common denominator (which is actually the same as least common multiple)

For example, in order to add 1/4 + 3/10, we need to find the smallest number that 4 and 10 will both divide into.
The least common multiple of 4 and 10 is 20, so we're going to find EQUIVALENT fractions for 1/4 and 3/10 so that both of these equivalent fractions have the same denominator.

KEY IDEA: We create equivalent fractions by multiplying (or dividing) the numerator and denominator by the SAME value.

Take 1/4 and multiply top and bottom by 5 to get 5/20.
So, 1/4 = 5/20

Now take 3/10 and multiply top and bottom by 2 to get 6/20.
So, 3/10 = 6/20

This means 1/4 + 3/10 = 5/20 + 6/20

Now that we have the same denominators, we can add the tops to get: 11/20


Now let's use the above findings to answer your other question: why 20x why not just 20 since 5 doesn't have an X?
So, you're asking why we can't rewrite all fractions with a denominator of 20 (instead of 20x)
Let's find out.

GIVEN EQUATION: 1/2x + 3/5 = 1 + 3/4x

Let's see what happens if we try to rewrite the three fractions (1/2x, 3/5, and 3/4x) with a denominator of 20

Let's start with 3/5
Multiply top and bottom by 4 to get 12/20.
So, 3/5 = 12/20 (so far so good)

Next we'll rewrite 1/2x with a denominator of 20
If we multiply top and bottom by 10, we get 10/20x, but we don't want a denominator of 20x. We want a denominator of 20.

So how do we accomplish this?
We need (2x)(something) = 20

We can't do it.

Does that help?


Great. It helped tremendously

Hi Brent, can I find common denominator on either side each? 10x on LHS and 4x on RHS? So I get (5+6x)/10x = (4x+3)/4x
Is this a valid way?
greenlight-admin's picture

That strategy (although a little longer) definitely works.
Once you have (5+6x)/10x = (4x+3)/4x, you can cross multiply to help solve.

Thanks Brent for confirmation.

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