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Understanding how to perform basic operations (i.e., addition, subtraction, multiplication, and division) on fractions is an essential skill for the GRE. If you’re already familiar with this topic, feel free to skip directly to the quiz at the end.

Just like with regular numbers, adding and subtracting fractions is essentially the same: subtracting is the same as adding a negative number. There are two key steps for fraction addition/subtraction:

- Adjust the fractions so they have the same denominators
- Add (or subtract) the numerators

Any fraction (or any number or any equation) can be adjusted into a different format by multiplying by $1$ expressed as a fraction. For example, what we if want to adjust the fraction $21 $ to have a denominator of $4$ instead of $2$? We could multiply it by $22 $:

$21 ∗22 =42 $

This adjustment doesn’t change the value of the number, since anything multiplied by $1$ is still the same quantity. It just changes the way that value is expressed.

Let’s consider a simple example:

What is $21 +43 $?

Since these two fractions have different denominators, we can adjust them, and then add them.

$x =21 +43 =(21 ∗1)+43 =(21 ∗22 )+43 =42 +43 =45 $

Now let’s try it with another simple example, this time including variables.

$3y/2+y/3−2y/12=1$

Solve for $y$.

Do you know the answer?

(spoiler)

$y=3/5$

First, adjust the fractions to have a common denominator. It’s often best to find the least common multiple (LCM) and use that, but any common multiple will work. In this case we’ll use 12, which means we need to adjust the first two terms.

First, let’s do $3y/2$. It has a denominator of $2$, so to get to $12$, we need to multiply by $6/6$.

$3y/2∗6/6=18y/12$

Next, let’s work on $y/3$. It has a denominator of $3$, so this time we need to multiply by $4/4$.

$y/3∗4/4=4y/12$

Now we can use these adjusted fractions in our equation to make it easier to work with. The denominators are identical, allowing us to perform simple math to evaluate the expression.

$3y/2+y/3−2y/1218y/12+4y/12−2y/12(18y+4y−2y)/1220y/1220yyy =1=1=1=1=12=12/20=3/5 $

Multiplication with fractions is straightforward, literally. You multiply straight forward across the two fractions, multiplying the numerators together, and multiplying the denominators together. For example:

$53 ∗72 =5∗73∗2 =356 $

As a reminder, you might see fractions written horizontally, but they’re still the same.

$53 =3/5$

Let’s work through a simple example that uses fraction multiplication.

$(3x/y_{2})(3x/4) =3$

What is the ratio of $x$ to $y$?

Know the answer?

(spoiler)

The ratio of $x$ to $y$ is $2$ to $1$

The ratio of $x$ to $y$ is the same as the answer to $x/y$. Here’s the step-by-step math:

$(3x/y_{2})(3x/4) (3x∗3x)/(y_{2}∗4) 9x_{2}/4y_{2} 3x/2yx/2yx/y =3=3=3=3=1=2 $

We’ve come to the answer that $x/y=2$ , which is really the same as $x/y=2/1$, which means the ratio of $x$ to $y$ is $2$ to $1$.

Dividing one fraction by another is very similar to multiplication. First, swap the numerator and denominator of the divisor (the second fraction), and then multiply. For example:

$72 ÷35 =72 ∗53 =356 $

Sometimes you’ll hear this explained as *multiplying by the reciprocal*. The reciprocal is the swapped fraction, e.g., the reciprocal of $35 $ is $53 $.

Here’s an example using fraction division:

The $n$th term of a sequence is $(n/(n+1))/(n/(n+1))$

What is the sum of the first 7 terms in the sequence?

If you haven’t read the chapter on sequences yet, don’t worry. Here’s another way to phrase the same question:

$f(n)=(n/(n+1))/(n/(n+1))$

What is the sum of $f(n)$ for each integer $n$ where $1<=n<=7$?

Don’t let the horizontal format, variables, and parentheses confuse you!

(spoiler)

Answer: 7

Start by finding the first term in the sequence, i.e., the value of the expression when $n=1$:

$f(1) =(n/(n+1))/(n/(n+1))=(1/(1+1))/(1/(1+1))=(1/2)/(1/2)=(1/2)∗(2/1)=2/2=1 $

That’s an interesting result… but if we look at the equation again carefully, it makes sense. Let’s look at it written vertically:

$f(n)=n/(n+1)n/(n+1) $

This fraction has the numerator divided by the same denominator, so it turns out that regardless of the value of $n$, the result will always be $1$. Therefore, the sum of the first 7 terms is $1+1+1+1+1+1+1=7$.

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