Lesson 2. Subtracting fractions

In the previous lesson, we have learned how to add fractions together. Let's first connect this to a formula you may have seen in high school. We have
$$\frac{A}{B} + \frac{C}{D} = \frac{AD}{BD} + \frac{BC}{BD} = \frac{AD + BC}{BD}.$$
That is, we have first multiplied the opposite bottoms ($D$ is the opposite bottom for $A/B$, and $B$ is the opposite bottom for $C/D$) up and down and then keep the same bottom $BD$ and added the two different tops. In many curricula, teachers describe this process as follows.

1. Multiply the bottoms.
2. Cross multiply: $A$ with $D$ and $C$ with $B$.
3. Then add the resulting fractions.

The method introduced in the previous lesson is equivalent to this cross multiplication method, and in fact, it teaches you why this cross multiplication is valid. You may use either way, but you will have a stronger understanding if you know both. (If you can forget one of them, you can use another!)

Subtraction of fractions. We can subtract a fraction from another similarly. In the math symbols above, just replace $+$ with $-$ everywhere:

$$\frac{A}{B} - \frac{C}{D} = \frac{AD}{BD} - \frac{BC}{BD} = \frac{AD - BC}{BD}.$$

Exercise. Compute $1/2 - 1/3$.

Solution. We have

$$\frac{1}{2} - \frac{1}{3} = \frac{1 \cdot 3}{2 \cdot 3} - \frac{2 \cdot 1}{2 \cdot 3} = \frac{3}{6} - \frac{2}{6} = \frac{3 - 2}{6} = \frac{1}{6}.$$

This is the answer.

Advice. The following exercise can be challenging. Try it first, and if it is too challenging, feel free to consult the solution. The regular GMAT questions have similar difficulty, so we need to step ourselves up a little bit. For this problem, you will have to recall that
$$B \cdot \frac{A}{B} = A.$$
That is, multiplying a fraction by its bottom cancels out the bottom. This is because $A/B$ means $A$ divided by $B$, and multiplication is the opposite operation to the division.

Exercise. If 

$$\frac{x}{2} - \frac{y}{3} = 1 = \frac{x}{3} - \frac{y}{2},$$

then what is $x + y$?

Solution. Note that we can split the given condition into the following two conditions:

1. $x/2 - y/3 = 1$,
2. $x/3 - y/2 = 1$.

The left-hand side of the first condition can be manipulated using our knowledge of subtraction of fractions:

$$\frac{x}{2} - \frac{y}{3} = \frac{x \cdot 3}{2 \cdot 3} - \frac{2 \cdot y}{2 \cdot 3} = \frac{3x}{6} - \frac{2y}{6} = \frac{3x - 2y}{6}.$$

Again, recall that this was the left-hand side of the first condition. What is the right-hand side of the first condition? It is $1$. Therefore, we must have

$$\frac{3x - 2y}{6} = 1.$$

Multiplying $6$ on the left-hand side will cancel $6$ at the bottom. Multiplying $6$ on the right-hand side will just give us $6$, so

$$3x - 2y = 6.$$

This is just another way to write the first condition. Working similarly, I believe in you that you will manipulate the second condition into

$$2x - 3y = 6.$$

So far, we have

1. $3x - 2y = 6$,
2. $2x - 3y = 6$.

Subtract the left-hand sides: $(3x - 2y) - (2x - 3y) = 3x - 2y - 2x + 3y = x + y$. Subtract the right-hand sides: $6 - 6 = 0$. This means that $x + y = 0$ is the answer to our exercise.