1 Million Means: 1 million in numerical is represented as 10,00,000. The Indian equivalent of a million is ten lakh rupees. It is not a...

1 Million Means: 1 Million in Rupees, Lakhs and Crores

June 5, 2024**Addition and Subtraction of Algebraic Expressions: **We have learned to use arithmetic symbols, mainly of the type \(0,1,2,3,4,5,…….\) These symbols are also referred to as numbers. We also used \(+\) sign to add, \(–\) sign to subtract, the \(\times \)sign to multiply and \( \div \) sign to perform arithmetic activities for the division. The new branch of mathematics called Algebra is being discussed on this page. Letters known as variables and numbers are used to form an expression using all basic mathematical operators known as the algebraic expression. We will also study the addition of these algebraic expressions and their subtraction.

Algebra is the branch that deals with mathematical symbols that do not have any firm values and are called variables and the rules to manipulate them. In our real-life problems, we often see specific values that keep on switching. But there is a need for a medium to represent these switching values.

Here in algebra, these values are often represented with \(x, y, z, p,\) or \(q,\) and these symbols are called variables. These symbols are handled through various arithmetic operations of addition, subtraction, multiplication, and division to find the values.

A combination of constants and variables connected by the fundamental mathematical operations of addition, subtraction, multiplication and division is called an algebraic expression. **Terms: **Various parts of an algebraic expression separated by the signs \(+\) or \(-\) are called the terms of the expression. The terms may contain a variable alone, a constant alone or a combination of variables or constants connected by multiplication or division.Examples: \(2,y,2x,\frac{3}{y},\) etc.

In the above algebraic expression, \(5\) and \(9\) are constants, \(x\) is variable, and the arithmetic operation of addition is performed.

Learn Everything About Algebraic Expressions

**Constants: **A term of the expression having no variable factor is called a constant term.**Coefficient: **In a term of an algebraic expression, any of the factors with the sign of the term is called the coefficient of the product of other factors.

Below are the different types of algebraic expressions.

- Monomial
- Binomial
- Trinomial
- Quadrinomial
- Polynomial

**Monomial**: An algebraic expression containing only one term is called a monomial.

Example: \(3,2x,5{x^2}y\) are all monomials**Binomial**: An algebraic expression containing two terms is called a binomial.

Example: \(x + 3,5 – 2x,{a^2} – 2abc\) are all binomials**Trinomial**: An algebraic expression containing three terms is called a trinomial.

Example: \(3x – y + 3,{x^2} + {y^2} + 2xy\) are all trinomials**Quadrinomial**: An algebraic expression containing four terms is called a quadrinomial.

Example: \(ab + bc + ca + abc,{a^2} + {b^2} + {c^2} + 5\) are all quadrinomials**Polynomial**: An algebraic expression containing one or more terms is called a polynomial.

Therefore, the monomial, binomial, trinomial and quadrinomial etc., can be termed as polynomials.

Like terms are the ones that are having the same algebraic factors. Unlike terms are the ones that are not having the algebraic factors.

Example: In the algebraic expression, \(5{a^2}b + 7a{b^2} – 3ab – 4b{a^2},\) we have \(5{a^2}b\) and \( – 4b{a^2}\) as like terms, whereas \(7a{b^2}\) and \( – 3ab\) are unlike terms.

We have now learnt about the algebraic expressions, the like and unlike terms in an algebraic expression. Let us study the fundamental operations of addition and subtraction of algebraic expressions since an algebraic expression may consist of like and unlike terms. So, the operations of addition and subtraction of algebraic expressions mean addition or subtraction of like terms.

**Addition of positive like terms: **To add the positive like terms, we follow the below steps:

1. Obtain all like terms.

2. Find the sum of the numerical coefficients of all terms.

3. Write the required sum as the like term whose numerical coefficient is the number obtained in the second step, and the variable factor is the same as the variable factors of the given like terms.

Example: Add \(4xy,5xy\) and \(9xy\)

Solution: The sum of the numerical coefficients of the given like terms is \(4+5+9=18.\)

Thus, the sum of the given like terms is another like term whose numerical coefficient is \(18.\)

Therefore, \(4xy + 5xy + 9xy = 18xy\)

**Addition of Negative Like Terms: **To add the negative like terms, we follow the below steps:

- Obtain all like terms.
- Obtain the sum of the numerical coefficients without negative signs of all like terms.
- Write an expression as a product of the number obtained in the second step, with all the variable coefficients preceded by a minus sign.

Example: Add \( – 7xy, – 3xy\) and \( – 8xy\)

Solution: The numerical coefficients (without the negative sign) of the given like terms are \(7, 3, 8.\)

Therefore, the sum of the numerical coefficients\(=7+3+8=18\)

So, the sum of the given like terms is another like term whose numerical coefficient is \(18.\)

Hence, \( – 7xy – 3xy – 8xy = – 18xy\)

If positive and negative like terms are involved, add the coefficients according to the general rule of addition of integers or rational numbers and continue like the above two methods.

To subtract an algebraic expression from another, we should change the signs (from \(+\) to \(-\) or from \(-\) to \(+\)) of all the terms of the expression to be subtracted and then the two expressions are added.

Example: Subtract \(-8a\) from \(-3a\)

Solution: \(-3a–8a=-3a+8a=5a\)

In adding algebraic expressions containing like and unlike terms, we collect different groups of like terms and find the sum of like terms in each group using the methods mentioned above. The collection of like terms can be done by any one of the following two methods:

**Horizontal Method****Column Method**

In this method, all expressions are written in a horizontal line and then the terms are arranged to collect all the groups of like terms and then added or subtracted as required.

Example:

1. Add \(3x + 2y\) and \(x + y\)

Solution: Adding \(3x + 2y\) and \(x + y\) using the horizontal method is shown below.

\((3x + 2y) + (x + y)\)

Grouping the like terms, we get

\((3x + x) + (2y + y)\)

\( \Rightarrow (3 + 1)x + (2 + 1)y\)

\( = 4x + 3y\)

So, \((3x + 2y) + (x + y) = 4x + 3y\)

2. Subtract: \({a^2} – 3ab\) from \(2{a^2} – 7ab\)

Solution: Subtracting \({a^2} – 3ab\) from \(2{a^2} – 7ab\) using the horizontal method is shown below.

\(\left( {2{a^2} – 7ab} \right) – \left( {{a^2} – 3ab} \right)\)

Grouping the like terms, we get

\(\left( {2{a^2} – {a^2}} \right) – 7ab + 3ab\)

\( = {a^2} – 4ab\)

Therefore, \(\left( {2{a^2} – 7ab} \right) – \left( {{a^2} – 3ab} \right) = {a^2} – 4ab\)

In this method, each expression is written in a separate row such that their like terms are arranged one below the other in a column. Then the addition or subtraction of terms is done column-wise.

**Example:**

- Add: \(x + y + 3\) and \(3x + 2y + 5\)

Solution: Adding \(x + y + 3\) and \(3x + 2y + 5\) using the column method is shown below.

Therefore, \((x + y + 3) + (3x + 2y + 5) = 4x + 3y + 8\)

(ii) Subtract: \({x^2} – 3xy + 7{y^2} – 2\) from \(6xy – 4{x^2} – {y^2} + 5\)

Solution: Subtracting \({x^2} – 3xy + 7{y^2} – 2\) from \(6xy – 4{x^2} – {y^2} + 5\) using the column method is shown below.

Therefore, \(\left( {6xy – 4{x^2} – {y^2} + 5} \right) – \left( {{x^2} – 3xy + 7{y^2} – 2} \right) = 9xy – 5{x^2} – 8{y^2} + 7\)

** Q.1. What should be added to** \({a^2} + 2ab + {b^2}\)

Therefore, required expression\( = \left( {4ab + {b^2}} \right) – \left( {{a^2} + 2ab + {b^2}} \right)\)

\( = 4ab + {b^2} – {a^2} – 2ab – {b^2}\)

\( = 4ab – 2ab + {b^2} – {b^2} – {a^2}\)

\( = 2ab – {a^2}\)

Therefore, the required algebraic expression is \(2ab – {a^2}.\)

** Q.2. What should be subtracted from** \({a^3} – 4{a^2} + 5a – 6\)

So, \(\left( {{a^3} – 4{a^2} + 5a – 6} \right) – P = {a^2} – 2a + 1\)

\( \Rightarrow \left( {{a^3} – 4{a^2} + 5a – 6} \right) – \left( {{a^2} – 2a + 1} \right) = P\)

\( \Rightarrow P = {a^3} – 4{a^2} + 5a – 6 – {a^2} + 2a – 1\)

\( \Rightarrow P = {a^3} – 4{a^2} – {a^2} + 5a + 2a – 6 – 1\)

\( \Rightarrow P = {a^3} – 5{a^2} + 7a – 7\)

Therefore, the required expression is \({a^3} – 5{a^2} + 7a – 7.\)

** Q.3. From the sum of** \(4{x^4} – 3{x^3} + 6{x^2},4{x^3} + 4x – 3\)

\(\left( {4{x^4} – 3{x^3} + 6{x^2}} \right) + \left( {4{x^3} + 4x – 3} \right) + \left( { – 3{x^4} – 5{x^2} + 2x} \right)\)

\( = 4{x^4} – 3{x^4} – 3{x^3} + 4{x^3} + 6{x^2} – 5{x^2} + 4x + 2x – 3\)

\( = {x^4} + {x^3} + {x^2} + 6x – 3\)

Now, we have to subtract \(5{x^4} – 7{x^3} – 3x + 4\) from \({x^4} + {x^3} + {x^2} + 6x – 3\)

\( \Rightarrow \left( {{x^4} + {x^3} + {x^2} + 6x – 3} \right) – \left( {5{x^4} – 7{x^3} – 3x + 4} \right)\)

\( = {x^4} + {x^3} + {x^2} + 6x – 3 – 5{x^4} + 7{x^3} + 3x – 4\)

\( = \left( {{x^4} – 5{x^4}} \right) + \left( {{x^3} + 7{x^3}} \right) + {x^2} + 6x + 3x – 3 – 4\)

\( = – 4{x^4} + 8{x^3} + {x^2} + 9x – 7\)

Therefore, the required expression is \( – 4{x^4} + 8{x^3} + {x^2} + 9x – 7\)

** Q.4. Add** \(7{x^2} – 4x + 5, – 3{x^2} + 2x – 1\)

Grouping the like terms, we get

\(7{x^2} – 3{x^2} + 5{x^2} – 4x + 2x – x + 5 – 1 + 9\)

\( = (7 – 3 + 5){x^2} + ( – 4 + 2 – 1)x + (5 – 1 + 9)\)

\( = 9{x^2} – 3x + 13\)

Therefore, \(\left( {7{x^2} – 4x + 5} \right) + \left( { – 3{x^2} + 2x – 1} \right) + \left( {5{x^2} – x + 9} \right) = 9{x^2} – 3x + 13\)

** Q.5. Subtract: **\(\left( { – 2{y^2} + \frac{1}{2}y – 3} \right)\)

\( = 7{y^2} – 2y + 10 + 2{y^2} – \frac{1}{2}y + 3\)

\( = (7 + 2){y^2} + \left( { – 2 – \frac{1}{2}} \right)y + 13\)

\( = 9{y^2} – \frac{5}{2}y + 13\)

Therefore, \(\left( {7{y^2} – 2y + 10} \right) – \left( { – 2{y^2} + \frac{1}{2}y – 3} \right) = 9{y^2} – \frac{5}{2}y + 13\)

In the above article, we have studied algebra, the definition of algebraic expressions, types of algebraic expressions and the meaning of like and unlike terms. Also, we have learned the definition of terms in algebraic expressions, constants in algebraic expression, variables in algebraic expressions and coefficients of algebraic expression. We have learnt the addition and subtraction of algebraic expressions and addition and subtraction using the horizontal and column method and solved some example problems.

*Q.1. How can we add algebraic expressions?*** Ans:** In adding algebraic expressions, we collect different groups of like terms and find the sum of like terms in each group. Note that the sum of several like terms is another like term whose coefficient is the sum of the coefficients of those like terms.

*Q.2. How can we subtract algebraic expressions?*** Ans:** To subtract an algebraic expression from another, we should change the signs (from \(+\) to \(-\) or from \(-\) to \(+\)) of all the terms of the expression to be subtracted, and then the two expressions are added.

*Q.3. Why do we change the sign in algebraic expressions while doing subtraction?*** Ans:** Let us take an example of two numbers, \(5\) and \(-4.\) Suppose if we have to subtract \(-4\) from \(5,\) we write it as \(5-(-4.)\) We know that the product of two positive signs is positive, the product of two negative signs are positive, and the product of two, unlike signs, are always negative. In the example: \(5-(-4)=5+4=9.\) If we don’t change the sign, then the above process results as \(5-4=1.\) That is a completely different result from the original. So, we have to change the sign while doing subtraction of algebraic expressions.

*Q.4. Can we add or subtract unlike terms in algebraic expressions?*** Ans:** No, we cannot add or subtract the unlike terms in algebra. For example, the sum of \(5x+3y\) is the same as \(5x+3y.\)

*Q.5. How do you add or subtract like terms?*** Ans:** Group all the like terms in the given expression. Then add or subtract their coefficients as required.

Example: \({\rm{(5 a + 2b) + (3 a – 9b) = (5a + 3a) + (2b – 9b) = 8a – 7b}}\)