Skip to main content

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.

Comments

  1. Kind_HelperMarch 1, 2020 at 4:47 PM

    Thank you so much for sharing this post on how to add and subtract fractions. You have explained it in a very understandable manner. I have been searching reading a lot of ways to add and subtract fractions and encountered only two ways. First is you used the LCM. This is best for smaller fractions. The other is to use butterfly method where you have to cross multiply the fractions. This method is best for bigger fractions and those that has no LCM. If you have doubt you can check your answers using fraction calculator with whole numbers. The step by step solution provided by the fraction calculator will let you guide if you miss a single step. Again thanks for sharing for this valuable information.

    ReplyDelete

Post a Comment

Popular posts from this blog

Lesson 5. Comparing fractions

One of many questions GMAT can ask you is how to compare two or more fractions (i.e., which fractions are bigger/smaller). We will discuss a very simple method for comparing fractions. How to compare two fractions . To compare two fractions, we need two steps. Step 1 . Make the same positive bottoms. For example, let's compare $1/13$ and $2/25$. To make the same bottom, you multiply the opposite bottom up and down. For $1/13$, the opposite bottom is $25$, we have $$\frac{1}{13} = \frac{1 \cdot 25}{13 \cdot 25} = \frac{25}{13 \cdot 25}.$$ Why didn't we compute $13 \cdot 25$? Just wait for it now. We will explain why after comparing $1/13$ and $2/25$. For $2/25$, the opposite bottom is $13$, so $$\frac{2}{25} = \frac{13 \cdot 2}{13 \cdot 25} = \frac{26}{13 \cdot 25}.$$ Step 2 . Compare the tops. We continue the example from Step 1. We have $$\frac{1}{13} = \frac{25}{13 \cdot 25} < \frac{26}{13 \cdot 25} = \frac{2}{25}.$$ This implies that $1/13$ is smaller than
Post a Comment
Read more

Lesson 7. Probability

Computing probability is simple, it is: SPECIFIC / TOTAL. What does it mean? Let's start with an example. Example . Suppose that you have a (six-sided) regular dice, whose sides contain numbers $1, 2, 3, 4, 5, 6$. If you throw the dice, the probability that you will get 6 can be computed as follows: TOTAL ($1, 2, 3, 4, 5, 6$) = $6$ sides; SPECIFIC (only $6$) = $1$ sides. Therefore, the probability that you get $6$ from throwing the dice once is equal to $1/6$.  Exercise . What is the probability that you get $3$ by throwing the same dice? Solution . TOTAL = $6$, as there are six sides for the dice, and SPECIFIC = $1$, because there is one side that has $3$. Thus, the probability that you get $3$ from throwing the dices once is equal to $1/6$. This is the answer. Let's spice up our example a little bit. Example . Suppose that you have two regular dices, each of which has six sides containing $1, 2, 3, 4, 5, 6$. If you throw both of them at once, wha
Post a Comment
Read more