Lesson Video: Addition and Subtraction of Algebraic Expressions Mathematics • First Year of Preparatory School

Video Transcript

In this video, we will learn how to add and subtract algebraic expressions by adding and subtracting like terms. The main skill we’ll need is the ability to recognize like terms. Like terms are terms whose variable and exponent are the same. If we have values like two and 10, they’re both whole numbers. So they are considered like terms and can be combined. However, two 𝑥 and 10 are not like terms. In this case, the two has a variable and the 10 does not. This means these are unlike terms.

What about two 𝑥 cubed and 10𝑥 to the fourth power? They do have the same variable. But they do not have the same variable taken to the same power, which means they cannot be combined and they’re not like terms. But if we had two 𝑥 cubed plus 10𝑥 cubed, since both of these terms have the same variable taken to the same exponent, they’re like terms and we combine them by adding their coefficients. Two 𝑥 cubed plus 10𝑥 cubed equals 12𝑥 cubed. If you’re wondering why this works, let’s look at this another way.

If we know that a pink block equals 𝑥 squared, a yellow block equals 𝑥, and a green block equals one and we wanted to rewrite this first set of blocks, instead of saying 𝑥 squared plus 𝑥 squared plus 𝑥 squared for the pink, we can write that as three 𝑥 squared. Since there are two yellow blocks, we have two 𝑥 and the two green blocks that are whole numbers are two. And if we follow the same procedure for the second set of blocks, we’ll have two 𝑥 squared plus 𝑥 plus two. And combining all of these together, we see that there are five pink blocks, which means there should be five 𝑥 squared. And this makes sense because three 𝑥 squared plus two 𝑥 squared would equal five 𝑥 squared. We combine their coefficients.

When it comes to the 𝑥-terms, the yellow blocks, we have three, two 𝑥 plus 𝑥 equals three 𝑥, and of course for our whole numbers, we have four green blocks, which each represent one, so four total. In this situation where each term represents a different color, you would never combine the green and the pink together. In the same way, we cannot combine an 𝑥 squared term with a whole number. Using this skill, we can look at some examples.

Simplify two 𝑠 plus one plus three 𝑠 plus two.

First, we can go ahead and copy down our expression. In order to simplify, we first need to consider if there’s anything inside the parentheses that can be simplified. Since two 𝑠 and one are unlike terms, we can’t add them together. The same thing is true for three 𝑠 and two; they can’t be added together. However, because this expression contains only addition and we can add in any order, we can remove the parentheses to see if there are any like terms that can be combined. Two 𝑠 and three 𝑠 both have the same variable and therefore can be combined by combining their coefficients. Two 𝑠 plus three 𝑠 will be equal to two plus three 𝑠. And we also know that we could add the two whole numbers together. One plus two is three, which makes the most simplified form of this expression, five 𝑠 plus three.

In our next example, we’ll first have to do some substitution. And then we’ll be able to add two expressions together.

Find 𝐴 plus 𝐵 given that 𝐴 equals five 𝑠 cubed minus three 𝑠 and 𝐵 equals negative six 𝑠 squared plus three 𝑠.

Our job is to find 𝐴 plus 𝐵, which means we plug in what we know 𝐴 is, five 𝑠 cubed minus three 𝑠. And then we substitute what we know 𝐵 is, negative six 𝑠 squared plus three 𝑠. In order to add 𝐴 plus 𝐵, we’ll have to see if we can find any like terms. We have an 𝑠 cubed term, two terms that just have the variable 𝑠, and one term that is 𝑠 squared. Since we’re only dealing with addition and subtraction here, we’re able to remove the parentheses. But again, the only values that can be added together are those that are like terms. That means they’re the ones with a variable taken to the same exponent.

In our case, we can combine negative three 𝑠 plus three 𝑠. We cannot combine five 𝑠 cubed and six 𝑠 squared. Even though they both have the variable 𝑠, their exponents are different. But how do we combine negative three 𝑠 plus three 𝑠? Well, negative three 𝑠 plus three 𝑠 equals zero. And that means, in its most simplified form, 𝐴 plus 𝐵 equals five 𝑠 cubed minus six 𝑠 squared.

And before we go any further and look at other examples, we should say that the same rules apply to subtraction of expressions. You can only subtract like terms. We could subtract three 𝑥 minus two 𝑥. And we do that by subtracting two from three, the coefficients, which is one 𝑥 or just 𝑥. However, we could not subtract two 𝑥 from three 𝑥 squared. They have the same variable but different exponents. And so we say that three 𝑥 squared minus two 𝑥 just equals three 𝑥 squared minus two 𝑥.

We’ll look at some other cases in our next example.

Find 𝐴 minus 𝐵 given that 𝐴 equals eight 𝑥 plus two and 𝐵 equals five 𝑥 minus one.

If we want to solve 𝐴 minus 𝐵, we plug in eight 𝑥 squared plus two for 𝐴 and five 𝑥 minus one for 𝐵. When we’re working with subtraction of terms, the parentheses are very important. This is because we are subtracting all of 𝐵 from 𝐴. And to do that, we have to distribute this subtraction across both terms for the expression in 𝐵. And that means we’re saying eight 𝑥 plus two minus five 𝑥 but plus one because we’re subtracting a negative one. And that means we’re adding. This is the step where if you’re not careful, you will get a sign mistake.

And so when you’re subtracting expressions, you need to be very careful to distribute the subtraction correctly. Once you do that, it’s a simple matter of combining like terms. We have two terms with an 𝑥-variable and two whole numbers. For our 𝑥-variable, we have eight 𝑥 minus five 𝑥. And that means we’ll subtract five from eight. Remember, the variable doesn’t change. We just do the subtraction from the coefficients. And then we have two plus one, which is three. Eight minus five is three. So we have three 𝑥 plus three. In this case, under these conditions, 𝐴 minus 𝐵 is equal to three 𝑥 plus three.

In our next example, we’ll deal with expressions that have three different variables.

Subtract six 𝑥 to the fifth power minus three 𝑦 cubed plus three 𝑧 squared from eight 𝑥 to the fifth power minus five 𝑦 cubed minus two 𝑧 squared.

To solve this problem, we need to think carefully about how to write out this subtraction. We’re subtracting six 𝑥 to the fifth power minus three 𝑦 cubed plus three 𝑧 squared from eight 𝑥 to the fifth power minus five 𝑦 cubed minus two 𝑧 squared. And this means that our first expression will be the eight 𝑥 to the fifth power minus five 𝑦 cubed minus two 𝑧 squared. And our second term will be six 𝑥 to the fifth power minus three 𝑦 cubed plus three 𝑧 squared. As we are subtracting the entire second expression from the first expression, we must put that second expression in parentheses and then distribute the subtraction across every term in the expression.

In this step, we pay close attention to the sign of every term. Six 𝑥 to the fifth power is positive. And we are going to subtract six 𝑥 to the fifth power. But the three 𝑦 cubed is negative. And if we want to subtract negative three 𝑦 cubed, we rewrite that as adding three 𝑦 cubed. The three 𝑧 squared is positive. We’re subtracting positive three 𝑧 squared, which we can rewrite as minus three 𝑧 squared. And in order to do any subtraction here, we need to see if we have like terms. Those are terms who have variables taken to the same exponent.

We have one term, that is, 𝑥 to the fifth power, and another 𝑥 to the fifth power term. We then have two terms with the variable 𝑦 cubed and two terms with the variable 𝑧 cubed. If you want, you can regroup so that the like terms are next to each other in the expression. As you’re doing this, pay close attention to the signs. And then we remember that, to combine like terms, we combine their coefficients. Eight 𝑥 to the fifth power minus six 𝑥 to the fifth power will be eight minus six 𝑥 to the fifth power, which is two 𝑥 to the fifth power.

For the second terms, we have negative five 𝑦 cubed plus three 𝑦 cubed. That means we need to combine negative five and positive three, which is negative two. And for our 𝑧 squared term, we’re combining the coefficients, negative two and negative three. Negative two minus three is negative five. Putting all this together, we get two 𝑥 to the fifth minus two 𝑦 cubed minus five 𝑧 squared.

Before we leave this question, let’s look at one other way you could solve, starting with our first expression. Instead of writing the second expression horizontally, you could vertically line the second expression up underneath the first. Notice that we’ve grouped the 𝑥 to the fifth terms, the 𝑦 cubed terms, and the 𝑧 squared terms. But again, our biggest challenge here is to make sure we don’t make any sign mistakes. For the first term, we’re saying eight 𝑥 to the fifth minus six 𝑥 to the fifth. So we subtract six from eight. And we get two 𝑥 to the fifth.

But our second set of like terms are not as straightforward. We have negative five 𝑦 cubed. And we are subtracting negative three 𝑦 cubed, negative five 𝑦 cubed plus three 𝑦 cubed. This is doing that distribution. Negative five plus three is negative two. And the variable is 𝑦 cubed. Then we have negative two 𝑧 squared minus positive three 𝑧 squared, which means we’ll have negative two 𝑧 squared minus three 𝑧 squared. Negative two minus three is negative five. And our variable is 𝑧 squared. And we see here that both methods will yield the same result.

In our final example, we’ll use the skills of adding algebraic expressions to solve for the perimeter of a right triangle.

Find the perimeter of the right triangle shown in the diagram.

We know that the perimeter of any shape is the distance all the way around the shape. In the case of this triangle, since all three sides have a different length, we need to add those three sides together. Side one has a measure of three 𝑥 plus side two, 𝑥 plus three, plus side three, two 𝑥 plus seven. Since we’re only dealing with addition, we can regroup the terms however we would like. And we want to group our like terms together. Three of our terms have an 𝑥-variable. And the other two terms are whole numbers.

We could rewrite the expression grouping the 𝑥-terms and the whole numbers. From there, for 𝑥-terms, we can combine these terms by combining their coefficients, which would be three plus one plus two, which is six. And to combine the whole numbers, of course, three plus seven is 10. Since we don’t have any information about what 𝑥 is, we can only say that the perimeter of this shape is equal to six 𝑥 plus 10.

Before we finish, let’s review the key points necessary for adding and subtracting algebraic expressions. First, we must recognize like terms. Like terms are terms whose variables and exponents are the same. To combine like terms, add or subtract their coefficients. When adding and subtracting two or more algebraic expressions, the only values that can be combined are the terms that are alike.