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Addition and Subtraction are the fundamental operations of Mathematics and can be applied to algebraic expressions. An algebraic expression is the combination of some constants, variables, and operators. The addition and subtraction of the algebraic expressions are quite similar to the addition and subtraction of integers only. Hence, when it comes to the case of algebraic expressions, one has to sort the like terms and unlike terms together. The algebraic expressions are formed from the combination of variables and constants. To understand the algebraic meaning of an expression, consider the expression 4xy + 7 which is formed from the variables x and y and constants 4 and 7.
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Key Terms: Algebraic Expressions, Terms, Variables, Coefficients, Like Terms, Unlike Terms, Polynomials, Constants, Horizontal Method, Column Method, Addition, Subtraction
What are Algebraic Expressions?
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Algebraic Expressions consists of variables and constants in which a variable can take any value, which means that its value is not fixed whereas the constant will have a fixed value. This indicates that the value of the algebraic expression value will change if the variable values are changed. We use alphabets such as x, y, l, m, ... etc. to denote variables.
Example: 9x + 23y + 14
Here 14 is the constant whereas x and y are variables of which 9 and 23 are the numerical coefficients respectively.
Algebraic Expression
Read More: Algebra and Its Branches
Before moving on to the addition and subtraction of the algebraic expressions, we must understand the concept of like terms and unlike terms:
- Like Terms
Those terms that contain the same variables which are raised to the same power are like terms. For example, 4xy and – 3xy are like terms as they have the same variables x and y with the same power 1. The numerical coefficients in like terms can vary. Only like terms can be added or subtracted with each other.
- Unlike Terms
Those terms that contain different variables that are not raised to the same power are unlike terms. For example, 4xy and – 3x are unlike terms as the first term has two variables x and y whereas the second term has only one variable x. Unlike Terms are not suitable for mathematical operations such as addition and subtraction.
The Following are the simple steps that help you to decide whether the given terms are like or unlike terms:
- Ignore the numerical coefficients. Concentrate on the algebraic part of the terms.
- Check the variables in the terms. They must be the same.
- The last step is to check the powers of each variable in the terms. If they are the same, the terms are like terms and if not, they are unlike terms.
Like Terms and Unlike Terms
Read More: Algebraic Operations on Complex Numbers
Addition of Algebraic Expressions
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The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms.
There are two different ways of solving the algebraic addition.
- Horizontal Method
In this method, one has to write all the expressions in a horizontal line and the terms are arranged to collect all the groups of like terms. These like terms are then added.
Consider the three algebraic terms: 5xy – 3x2 – 12y +5x; xy – 3x – 12yz + 5x3 and
y – 6x2 – zy + 5x3.
- Step 1: First add the three terms simply by putting the addition sign between them.
(5xy – 3x2 – 12y +5x) + (xy – 3x – 12yz + 5x3) + (y – 6x2 – zy + 5x3)
- Step 2: Next step is to open the brackets and multiply the signs.
5xy – 3x2 – 12y + 5x + xy – 3x – 12yz + 5x3 + y – 6x2 – zy + 5x3
- Step 3: Now, combine the like terms together.
(5xy + xy) + (-3x2 – 6x2) + (-12y + y) + (5x – 3x) + (-12yz – yz) + (5x3 + 5x3)
- Step 4: Add the coefficients while keeping the variables and the exponents on the variables intact.
6xy – 9x2 – 11y + 2x – 13yz + 10x3
Horizontal Method of Addition of Algebraic Expressions
Read More: Multiplication and Division of Integers
- Column Method
In this method, one needs to write each expression in a separate row in a way such that their like terms are arranged one below the other in a column. Then the terms are added column-wise.
Consider the addition of three expressions in columns
6a + 8b - 7c
- 4a + 2b + c
a - 3b - 2c
Here is how to add them:
6a + 8b - 7c
-4a + 2b + c
a - 3b - 2c
____________
3a + 7b - 8c
_____________
Read More: Standard Algebraic Identities
Subtraction of Algebraic Expressions
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Similar to the addition of the algebraic expressions, we can subtract some algebraic expressions in two different methods. Let’s work out these two methods along with the example.
- Horizontal Method
Consider the three terms as y3 + 3x2y – 6x2 – 6zy + 7x3 and x2y – 2x2 – zy + 5 and –3x2 + 3x3
- Step 1: Subtract the terms by using the subtraction sign.
(y3 + 3x2y – 6x2 – 6zy + 7x3) – (x2y – 2x2 – zy + 5) – (–3x2 + 3x3)
- Step 2: Open the brackets and then multiply the signs.
y3 + 3x2y – 6x2 – 6zy + 7x3 – x2y + 2x2 + zy – 5 + 3x2 – 3x3
- Step 3: Next step is to combine the like terms together.
y3 + ( 3x2y – x2y ) + (-6x2 + 2x2 + 3x2 ) + (-6zy + zy ) + (7x3 – 3x3 ) + (-5)
- Step 4: Add the coefficients while keeping the variables and exponents on the variables the same.
y3 + 2x2y – x2 – 5zy + 4x3 – 5
- Column Method
Arrange the expression in column-wise order and subtract the upper expression from the below one as follows. One thing to keep in mind is that the sign of each term that is to be subtracted is reversed and then the resulting expression is added normally.
Subtract 4a + 5b - 3c from 6a - 3b + c
6a - 3b + c
4a + 5b - 3c
(-) (-) (-)
_____________
2a - 8b + 4c
____________
Column Method of Subtraction of Algebraic Expressions
Read More: Algebra Formula
Things to Remember
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- The subtraction and addition of the algebraic terms are the same as the mathematical operations between the numbers.
- One must keep the concept of like terms and unlike terms before adding or subtracting any algebraic expression.
- Only like terms can be added or subtracted with each other.
- Unlike Terms can not be added or subtracted in any algebraic expression.
- There are two methods for both addition and subtraction of algebraic expressions namely the Horizontal Method and Column Method.
- In the horizontal method, all the expressions are written in a horizontal line and the terms are arranged to collect all the groups of like terms which will be added or subtracted.
- In the column method, each expression is in a separate row in a way such that their like terms are arranged one below the other in a column which is then added or subtracted.
- The important point to be noted here is that in Colum Method for Subtraction, the sign of each term that is to be subtracted is reversed and then the resulting expression is added normally.
Sample Questions
Ques. What is the coefficient of x2 when you add 2x4 + 5x3 and 6x4 − 7x3 + 5? (3 Marks)
Ans. The sum of 2x4 + 5x3 and 6x4 − 7x3 + 5 = (2x4 + 5x3) + (6x4 − 7x3+ 5)
= 2x4 + 5x3 + 6x4− 7x3+ 5.
Collecting all like terms = 2x4 + 6x4 + 5x3 − 7x3 + 5 = 8x4 − 2x3+ 5.
Therefore, the sum is 8x4 − 2x3+ 5 and the coefficient of x2 is 0. Since there is no expression with variable x2 in the sum.
Hence the coefficient of the term x2 is 0.
Ques. What is the value of (x+3y)2+ (x+3y)2 is? (3 Marks)
Ans. This is the simple addition of the terms after expanding the square terms.
(x+3y)2 = x2 + 9y2 + 6xy.
Hence adding the similar terms, we get
x2 + 9y2 + 6xy + x2 + 9y2 + 6xy = 2x2+ 18y2+ 12xy.
Therefore, the required answer is 2x2+ 18y2+ 12xy.
Ques. Roses and marigolds are planted in square plots in a garden. The length of the square plot in which marigolds are planted is 3 meters greater than that of the length of the square plot in which roses are planted. Calculate how much bigger in the area is the marigold plot than the rose plot? (5 Marks)
Ans. Let the length of the side of the marigold plot = x m
Length of the side of the plot =( x+2) m
Area of square is given by = side × side = (side)2
Area of Marigold Plot = x × x = x2
Area of Rose Plot = (x+2) × (x+2)=( x+2)2 = x2 + 22 + 2 × x × 2 = x2 + 4 + 2x [since, (a + b )2 = a2 + b2+ 2ab]
Difference in both the areas of square plots is = x2 + 4 + 2x - x2 = 2x + 4 = 2(x + 2) m
Thus, it is clear that the Rose plot is bigger in terms of area than the marigold plot by 2(x+2) m.
Hence, the rose plot is 2 (x + 2) m bigger than the marigold plot.
Ques. Simplify the expression by collecting together the like terms: (3 Marks)
12m2 – 9m + 5m – 4m2 – 7m + 10
Ans. By rearranging the terms, we have
12m2 – 4m2 + 5m – 9m – 7m + 10
= (12 – 4) m2 + (5 – 9 – 7) m + 10
= 8m2 + (– 4 – 7) m + 10
= 8m2 + (–11) m + 10
= 8m2 –11m + 10
Hence the final simplified expression is 8m2 –11m + 10
Ques. What will be the value of the following expressions with n = – 2? (5 Marks)
(i) 5n – 2
(ii) 5n2 + 5n – 2
(iii) n^3 + 5n2 + 5n – 2
Ans. (i) Putting the value as n = – 2, in the 5n – 2, we get,
5 (– 2) – 2 = – 10 – 2 = – 12
(ii) In 5n2 + 5n – 2, we have,
for n = –2, we have 5n – 2 = –12
and 5n2 = 5 × (– 2)2 = 5 × 4 = 20
Now Combining, 5n2 + 5n – 2 = 20 – 12 = 8
(iii) Now, for n = – 2,
5n2 + 5n – 2 = 8 and
n^3 = (–2)^3 = (–2) × (–2) × (–2) = – 8
Combining,
n^3 + 5n2 + 5n – 2 = – 8 + 8 = 0
Ques. Subtract 24ab – 10b – 18a from 30ab + 12b + 14a ? (3 Marks)
Ans. from the above steps for subtraction of two numbers, we can calculate it as :
30ab + 12b + 14a – (24ab – 10b – 18a)
Merging the like and dislike terms
= 30ab + 12b + 14a – 24ab + 10b + 18a
Simplifying by basic addition.
= 30ab – 24ab + 12b + 10b + 14a + 18a
= 6ab + 22b + 32a
Hence the final required answer is 6ab + 22b + 32a.
Ques. Sarita has some marbles. Ameen has 10 more. Ravi says that he has 3 more marbles than the number of marbles Sarita and Ameen together have. find the number of marbles that Ravi has? (3 Marks)
Ans. As it is not given the marbles that Saritha have,
Let’s consider them to be x.
Ameen says that he had all together greater than 10 so the number of marbles with Ameen becomes x + 10
Since Ravi says that he has 3 more marbles than what Saritha has, then he would have x + 3 + x + 10.
= 2x + 13.
Ques. Simplify the following: (5 Marks)
(i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
Ans. (i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
= a2b2 – a2c2) + b2c2 – b2a2) + c2a2 – c2b2)
= 0
After simplification, the answer is 0.
(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
= x3 – 3x2y2 – xy3 + 2x2y2 – xy3 + 5x3
= x3 + 5x3 – 3x2y2 + 2x2y2 – xy3 – xy3
= 6x3 – x2y2 – 2xy3
After simplification, the answer is 6x3 – x2y2 – 2xy3.
Ques. Subtract: 3x2 – 5x + 7 from 5x2 – 7x + 9 (3 Marks)
Ans. (5x2 – 7x + 9) – (3x2 – 5x + 7)
= 5x2 – 7x + 9 – 3x2 + 5x – 7
= 5x2 – 3x2 + 5x – 7x + 9 – 7
= 2x2 – 2x + 2.
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