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Since an algebraic expression may consist of like and unlike terms. So, the operations of addition of algebraic expressions mean addition of like terms.

#### Addition of Positive Like terms:

To add several positive like terms, we proceed as follows:

Step 1) Obtain all like terms.

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

Step 3) Write the required sum as a like term whose numerical coefficient is the numerical obtained in step (2) and literal factor is same as the literal factors of the given like terms.

**Example 1:** Number of apples with ajay is equal to 7 times the apples with vijay. What is the total number of apples both have together?**Solution:** Number of apples with vijay is not given in the question, we shall take the number as ‘x’

Ajay has 7 times as that of vijay i.e., 7 x x = 7x

To find total number of apples, we have to add x and 7x.

Therefore, the total number of apples = x + 7x = (1 + 7)x = 8x (distributive law)

**Example 2:** Add 12pq, 5pq and 6pq.**Solution:** The sum of the numerical coefficients of the given like terms is 12 + 5 + 6 = 23.

Thus, the sum of the given like terms is another like term whose numerical coefficient is 23.

Hence, 12pq + 5pq + 6pq = 23pq.

The sum of the given like terms can also be obtained by using the **distributive property of multiplication over addition** as discussed below:

12pq + 5pq + 6pq = (12 + 5 + 6)pq = 23pq.

**Example 3:** Add 7x\(y^2\), 11x\(y^2\) and 4x\(y^2\).**Solution:** The sum of the numerical coefficients of the given like terms is 7 + 11 + 4 = 22.

So, the sum of the given like terms is another like term whose numerical coefficient is 22.

Hence, 7x\(y^2\) + 11x\(y^2\) + 4x\(y^2\) = 22x\(y^2\).

The sum of the given like terms can also be obtained by using the distributive property of multiplication over addition as discussed below:

7x\(y^2\) + 11x\(y^2\) + 4x\(y^2\) = (7 + 11 + 4)7x\(y^2\) + 11x\(y^2\) + 4x\(y^2\)

= 22x\(y^2\)

#### Addition of Negative Like terms:

To add several negative like terms, we proceed as follows:

Step 1) Obtain all like terms.

Step 2) Find the sum of the numerical coefficients( without negative sign ) of all the terms.

Step 3) Write the required sum as a like term whose numerical coefficient is the numerical obtained in step (2) and literal factor is same as the literal factors of the given like terms preceded by a minus sign.

**Example 1:** Add -7pq, -6pq and -14pq.**Solution:** The numerical coefficients ( without negative sign ) of the given like terms are -7, -6 and -14.

Therefore, sum of the numerical coefficients of the given like terms is 7 + 6 + 14 = 27.

Thus, the sum of the given like terms is another like term whose numerical coefficient is 27.

Hence, -7pq – 6pq – 14pq = -27pq.

Using distributive property of multiplication over addition we have,

-7pq – 6pq – 14pq = (-7 – 6 – 14)pq = -27pq.

**Example 2:** Add -4x\(y^2\), -7x\(y^2\), -10x\(y^2\) and -3x\(y^2\).**Solution:** The sum of the numerical coefficients( without negative sign ) of the given like terms is 4 + 7 + 10 + 3 = 24.

So, the sum of the given like terms is another like term whose numerical coefficient is 24.

Hence, -4x\(y^2\) – 7x\(y^2\) – 10x\(y^2\) – 3x\(y^2\) = -24x\(y^2\).

The sum of the given like terms can also be obtained by using the distributive property of multiplication over addition as discussed below:

-4x\(y^2\) – 7x\(y^2\) – 10x\(y^2\) – 3x\(y^2\) = (-4 – 7 – 10 – 3)x\(y^2\)

= -24x\(y^2\)

#### Addition of Positive and Negative Like terms:

To add positive and negative like terms, we proceed as follows:

Step 1) Collect all positive like terms and find their sum.

Step 2) Collect all negative like terms and find their sum.

Step 3) Obtain the numerical coefficients( without negative sign ) of like terms obtained in Steps (1) and (2).

Step 4) Subtract the numerical coefficient in Step (2) from the numerical coefficient in Step (1). Write the answer as a product of this number and all the literal coefficients.

**Example 1:** Add 7x\(y^2\), 6x\(y^2\) and —2x\(y^2\).**Solution:** We have,

7x\(y^2\) + 6x\(y^2\) – 2x\(y^2\)

= (7x\(y^2\) + 6x\(y^2\)) – 2x\(y^2\) { Collecting +ve and -ve like terms together }

= 12x\(y^2\) – 2x\(y^2\)

= 10x\(y^2\)

Using distributive property of multiplication over addition: 7x\(y^2\) + 6x\(y^2\) – 2x\(y^2\) = (7 + 6 – 2)x\(y^2\) = 10x\(y^2\).

**Example 2:** Add 4ab, —7ab, -10ab and 3ab.**Solution:** We have,

4ab – 7ab – 10ab – 3ab

= 4ab + 3ab – 7ab – 10ab { Collecting +ve and -ve like terms together }

= 7ab – l7ab { Since, 4ab + 3ab = 7ab and -7ab – 10ab = -17ab }

= -10ab { Since, 7 – 17 = –10 }

Using distributive property of multiplication over addition: 4ab – 7ab – 10ab – 3ab = (4 – 7 – 10 + 3)ab = -10ab.

**Example 3:** Add 9\(x^2\) – 6x + 3, -5\(x^2\) – 3x + 8, 2\(x^2\) + 13x – 1.

**Solution:** Required sum = (9\(x^2\) – 6x + 3) + (-5\(x^2\) – 3x + 8) + (2\(x^2\) + 13x – 1)

By collecting like terms and adding them we have,

(9\(x^2\) – 5\(x^2\) + 2\(x^2\)) + (-6x – 3x + 13x) + (3 + 8 – 1)

By adding the coefficients of like terms we have,

(9 – 5 + 2)\(x^2\) + (-6 – 3 + 13)x + (3 + 8 – 1)

= 6\(x^2\) + 4x + 10.

**Example 4:** Add -16\(x^2\) + 8x + 27, 11\(x^2\) – 12x – 17, 7\(x^2\) – 19x – 21.

**Solution:** Required sum = (-16\(x^2\) + 8x + 27) + (11\(x^2\) – 12x – 17) + (7\(x^2\) – 19x – 21)

By collecting like terms and adding them we have,

(-16\(x^2\) + 11\(x^2\) + 7\(x^2\)) + (8x – 12x – 19x) + (27 – 17 – 21)

By adding the coefficients of like terms we have,

(-16 + 11 + 7)\(x^2\) + (8 – 12 – 19)x + (27 – 17 – 21)

= 2\(x^2\) – 23x – 11.

**Example 5:** Add (3\(x^2\) – \(\frac{1}{5}\)x + \(\frac{7}{3}\)), (-\(\frac{1}{4}\)\(x^2\) + \(\frac{1}{3}\)x – \(\frac{1}{6}\)), (-2\(x^2\) – \(\frac{1}{2}\)x + 5).

**Solution:** Required sum = (3\(x^2\) – \(\frac{1}{5}\)x + \(\frac{7}{3}\)) + (-\(\frac{1}{4}\)\(x^2\) + \(\frac{1}{3}\)x – \(\frac{1}{6}\)) + (-2\(x^2\) – \(\frac{1}{2}\)x + 5)

By collecting like terms and adding them we have,

(3\(x^2\) – \(\frac{1}{4}\)\(x^2\) – 2\(x^2\)) + (-\(\frac{1}{5}\)x + \(\frac{1}{3}\)x – \(\frac{1}{2}\)x) + (\(\frac{7}{3}\) – \(\frac{1}{6}\) + 5)

By adding the coefficients of like terms we have,

(3 – \(\frac{1}{4}\) – 2)\(x^2\) + (-\(\frac{1}{5}\) + \(\frac{1}{3}\) – \(\frac{1}{2}\))x + (\(\frac{7}{3}\) – \(\frac{1}{6}\) + 5)

= (\(\frac{12 – 1 – 8}{4}\))\(x^2\) + (\(\frac{-6 + 10 – 15}{30}\))x + (\(\frac{14 – 1 + 30}{6}\))

= \(\frac{3}{4}\)\(x^2\) – \(\frac{11}{30}\)x + \(\frac{43}{6}\).

#### Addition of Algebraic Expressions with Like and Unlike terms:

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 by the methods of addition of algebraic expressions containing like terms. The collection of like terms can be done by any one of the following two methods:

1) **Row or Horizontal Method**

2) **Column or Vertical Method**