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Lesson Explainer: Addition and Subtraction of Algebraic Expressions Mathematics • 7th Grade

In this explainer, we will learn how to add and subtract algebraic expressions by adding and subtracting like terms.

We first recall that we can add or subtract monomial expressions (products of constants and variables raised to integer exponents) with the same variables raised to the same exponents (called like terms) by adding or subtracting their coefficients. For example, if we want to simplify 5π‘₯+3π‘₯, we can note that these are like terms. In particular, we can take out the factor of π‘₯ to get 5π‘₯+3π‘₯=(5+3)π‘₯=8π‘₯.

We can apply this process with any number of terms. For example, we can simplify 7π‘₯βˆ’2π‘₯+3+6π‘₯βˆ’2π‘₯ by rearranging the terms into descending exponents: 7π‘₯βˆ’2π‘₯+3+6π‘₯βˆ’2π‘₯=7π‘₯βˆ’2π‘₯βˆ’2π‘₯+6π‘₯+3.

Then, we combine like terms: 7π‘₯βˆ’2π‘₯βˆ’2π‘₯+6π‘₯+3=(7βˆ’2)π‘₯+(βˆ’2+6)π‘₯+3=5π‘₯+4π‘₯+3.

We can extend this idea to simplifying the sum or difference of algebraic expressions by collecting like terms and simplifying. For example, imagine we are asked to add 3π‘₯+1 to βˆ’7π‘₯+2. We can represent this as (3π‘₯+1)+(βˆ’7π‘₯+2). We note that we can rearrange the sum of these terms to give (3π‘₯+1)+(βˆ’7π‘₯+2)=(3π‘₯βˆ’7π‘₯)+(1+2).

Now, we combine like terms: (3π‘₯βˆ’7π‘₯)+(1+2)=(3βˆ’7)π‘₯+3=βˆ’4π‘₯+3.

This is sometimes referred to as adding the expressions β€œhorizontally” or β€œthe horizontal method,” since all the expressions are written on the same line. However, this is not the only method we can use. We can also simplify these expressions vertically.

For example, imagine we want to simplify ο€Ή3π‘₯𝑦+5π‘§ο…βˆ’ο€Ή2π‘₯π‘¦βˆ’π‘§+1ο…οŠ¨οŠ¨. We can write this subtraction vertically, where we align like terms in the same column as shown: 3π‘₯𝑦+5𝑧+0βˆ’(2π‘₯π‘¦βˆ’π‘§+1)

Observe that we include the term 0 to help keep track of the terms in each column. We can now subtract the terms in each column. We have 3π‘₯π‘¦βˆ’2π‘₯𝑦=π‘₯π‘¦οŠ¨οŠ¨οŠ¨, 5π‘§βˆ’(βˆ’π‘§)=6𝑧, and 0βˆ’1=βˆ’1. We can write this below the bar as follows: 3π‘₯𝑦+5𝑧+0βˆ’(2π‘₯π‘¦βˆ’π‘§+1)π‘₯𝑦+6π‘§βˆ’1

These are known as the horizontal and column methods for adding and subtracting algebraic expressions respectively.

Let’s now see another example of how to apply the horizontal method to determine the simplified form of the addition of two algebraic expressions.

Example 1: Adding Two Algebraic Expressions

Simplify (2𝑠+1)+(3𝑠+2).


We recall we can add algebraic expressions by combining like terms. We first reorder the addition so that like terms are added together: (2𝑠+1)+(3𝑠+2)=(2𝑠+3𝑠)+(1+2).

We can then add like terms by adding their coefficients. This gives (2𝑠+3𝑠)+(1+2)=(2+3)𝑠+(1+2)=5𝑠+3.

In our next example, we will find the difference between two algebraic expressions by subtracting them vertically.

Example 2: Subtracting Two Algebraic Expressions by Writing them Vertically

Using the vertical method, subtract the following algebraic expressions: 2π‘₯+5𝑦+7βˆ’(3π‘₯βˆ’2π‘¦βˆ’1)


We first recall that we can subtract algebraic expressions by subtracting the coefficients of the like termsβ€”that is, the terms with the same variables and with the same exponents on these variables. We see that each column contains like terms, so we just need to subtract their coefficients.

We have 2π‘₯βˆ’3π‘₯=(2βˆ’3)π‘₯=βˆ’π‘₯,5π‘¦βˆ’(βˆ’2𝑦)=(5+2)𝑦=7𝑦,7βˆ’(βˆ’1)=7+1=8.

Writing this in a column gives the following: 2π‘₯+5𝑦+7βˆ’(3π‘₯βˆ’2π‘¦βˆ’1)βˆ’π‘₯+7𝑦+8

In our third example, we will solve a real-world problem that involves simplifying the difference of algebraic expressions.

Example 3: Subtracting Expressions to Solve a Word Problem

Sameh had 3π‘₯+6 coins and gave 2π‘₯+3 coins to Bassem. How many coins does Sameh have left?


The number of coins that Sameh has left will be the number of coins he has initially minus the number of coins he gives away. Since he starts with 3π‘₯+6 coins and gives 2π‘₯+3 to Bassem, he must have (3π‘₯+6)βˆ’(2π‘₯+3) coins left.

We can evaluate this difference by subtracting the coefficients of the like terms. We have (3π‘₯+6)βˆ’(2π‘₯+3)=(3π‘₯βˆ’2π‘₯)+(6βˆ’3)=(3βˆ’2)π‘₯+(6βˆ’3)=π‘₯+3.

Hence, Sameh has π‘₯+3 coins left.

In our next example, we will solve a real-world problem that involves both the addition and subtraction of algebraic expressions.

Example 4: Adding and Subtracting Expressions to Solve a Real-World Problem

Rania has 2π‘₯+7dollars. She gives 3π‘¦βˆ’2dollars to Karim and then receives π‘₯βˆ’π‘¦dollars from Adel. Write a simplified expression for the remaining amount of money, in dollars, that Rania has.


We first note that the amount of dollars that Rania has will be the amount she starts with minus the amount she gives away plus the amount of money she receives. We can write this as (2π‘₯+7)βˆ’(3π‘¦βˆ’2)+(π‘₯βˆ’π‘¦). We can simplify this expression by combining like terms. We then have (2π‘₯+7)βˆ’(3π‘¦βˆ’2)+(π‘₯βˆ’π‘¦)=(2π‘₯+π‘₯)+(βˆ’3𝑦+(βˆ’π‘¦))+(7βˆ’(βˆ’2))=3π‘₯βˆ’4𝑦+9.

In our next example, we will determine the areas of two different geometric constructs.

Example 5: Finding the Areas of Geometric Shapes

Consider three rectangles whose areas are represented by the expressions below.

  1. All three rectangles are put together such that none of them overlap. Write an expression to represent the area of the new combined shape.
  2. If shape 𝐴 is removed, and instead it sticks to a fourth rectangle, 𝐷, such that the shapes 𝐴 and 𝐷 do not overlap, the combined area of 𝐴 and 𝐷 is 16π‘¦π‘§βˆ’5.
    Write an expression to represent the area of rectangle 𝐷.


Part 1

Since all three rectangles do not overlap, the area of the combined shape will be the sum of their areas. We can add all of the areas together by reordering the terms so that we can combine like terms. We have (13π‘₯βˆ’7𝑦𝑧)+(2π‘₯+4𝑦𝑧)+(βˆ’4π‘₯+9𝑦𝑧)=(13π‘₯+2π‘₯βˆ’4π‘₯)+(βˆ’7𝑦𝑧+4𝑦𝑧+9𝑦𝑧).

Now, we combine the like terms by adding and subtracting their coefficients; this gives (13π‘₯+2π‘₯βˆ’4π‘₯)+(βˆ’7𝑦𝑧+4𝑦𝑧+9𝑦𝑧)=(13+2βˆ’4)π‘₯+(βˆ’7+4+9)𝑦𝑧=11π‘₯+6𝑦𝑧.

Part 2

Since shapes 𝐴 and 𝐷 do not overlap, the sum of their areas must equal 16π‘¦π‘§βˆ’5. If we say that shape 𝐷 has an area of 𝑑 square units, then we have that 13π‘₯βˆ’7𝑦𝑧+𝑑=16π‘¦π‘§βˆ’5.

We can solve for 𝑑 by subtracting the area of shape 𝐴 from the area of the combined shape. Hence, 𝑑=(16π‘¦π‘§βˆ’5)βˆ’(13π‘₯βˆ’7𝑦𝑧).

We reorder the terms as follows: 𝑑=βˆ’13π‘₯+(16π‘¦π‘§βˆ’(βˆ’7𝑦𝑧))βˆ’5.

Finally, we combine like terms. This gives 𝑑=βˆ’13π‘₯+(16βˆ’(βˆ’7))π‘¦π‘§βˆ’5=βˆ’13π‘₯+23π‘¦π‘§βˆ’5.

In our final example, we will use these concepts and our knowledge of simplifying monomials to find an expression for the surface area of a rectangular prism using its dimensions or side lengths.

Example 6: Finding an Expression for the Surface Area of a Rectangular Prism

Find the surface area of a rectangular prism with side lengths π‘₯, 11𝑦, and 15𝑧.


We can sketch the rectangular prism with these dimensions as follows.

We then recall that the surface area of a polyhedron is the sum of the areas of each of its faces. Since a rectangular prism has rectangular faces, these areas will be given by the products of pairs of its side lengths. We can find expressions for each of these areas by multiplying the given lengths. We have π‘₯Γ—11𝑦=(1Γ—11)π‘₯𝑦=11π‘₯𝑦,π‘₯Γ—15𝑧=(1Γ—15)π‘₯𝑧=15π‘₯𝑧,(11𝑦)Γ—(15𝑧)=(11Γ—15)𝑦𝑧=165𝑦𝑧.

We can add these areas to the diagram as shown.

Since opposite faces on a rectangular prism have equal areas, the surface area of this rectangular prism will be the sum of double each of these expressions. Therefore, surfacearea=2Γ—(11π‘₯𝑦)+2Γ—(165𝑦𝑧)+2Γ—(15π‘₯𝑧)=(2Γ—11)π‘₯𝑦+(2Γ—165)𝑦𝑧+(2Γ—15)π‘₯𝑧=22π‘₯𝑦+330𝑦𝑧+30π‘₯𝑧.

Hence, the surface area of the rectangular prism is given by the expression 22π‘₯𝑦+330𝑦𝑧+30π‘₯𝑧.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We can add and subtract algebraic expressions by combining like terms. In particular, we add and subtract the coefficients of the like terms.
  • We can evaluate these additions or subtractions horizontally, by writing all expressions in a horizontal line, then rearranging and collecting like terms, or vertically, by writing each expression in a separate row such that like terms are in the same column. These are known as the horizontal and vertical/column methods respectively.

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