 # Worksheet on Terms | Like and Unlike Terms in Algebraic Expressions Worksheets

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## Free Like and Unlike Terms Worksheet with Answers

Question 1.
(i) 9xy and xy
(ii) 3mnr, mnr, and 5mnr
(iii) 4a, 2a and 3b
(iv) m and 2n
(v) 6xy, 2yx and 8yz
(vi) 2m, 4m, and 5mn

Solution:

(i) Given that 9xy and xy.
From the given numbers, 9xy and xy are the like terms as they consist of the same variables, and the power of the like terms remains the same.
Now, add the sum of the coefficients of like terms.
9 + 1[xy] = 10xy

Therefore, 9xy + xy = 10xy.

(ii) Given that 3mnr, mnr, and 5mnr.
From the given numbers, 3mnr, mnr, and 5mnr are the like terms.
Now, add the sum of the coefficients of like terms.
{3 + 1 + 5}[mnr] = 9mnr

Therefore, 3mnr + mnr + 5mnr = 9mnr.

(iii) Given that 4a, 2a and 3b.
From the given numbers, 4a and 2a are the like terms.
Now, add the sum of the coefficients of like terms.
{4 + 2}[a] = 6a

Therefore, 4a + 2a + 3b = 6a + 3b.

(iv) Given that m and 2n.
From the given numbers, m and 2n both are unlike terms.

Therefore, the answer is m + 2n.

(v) Given that 6xy, 2yx and 8yz.
2yx can also write as 2xy.
From the given numbers, 6xy and 2xy are the like terms.
Now, add the sum of the coefficients of like terms.
{6 + 2}[xy] = 8xy

Therefore, 6xy + 2xy + 8yz = 8xy + 8yz.

(vi) Given that 2m, 4m, and 5mn.
From the given numbers, 2m and 4m are the like terms.
Now, add the sum of the coefficients of like terms.
{2 + 4}[m] = 6m

Therefore, 2m + 4m + 5mn = 6m + 5mn.

Question 2. If A = p2 + 2pq + q2, B = 2p2 – 3pq + 5q2 and C = 4p2 – pq – 7q2, find A + B + C.

Solution:

Given that A = p2 + 2pq + q2, B = 2p2 – 3pq + 5q2 and C = 4p2 – pq – 7q2.
Now, add A + B + C = (p2 + 2pq + q2) + (2p2 – 3pq + 5q2) + (4p2 – pq – 7q2)
From the above equation, p2, 2p2, and 4p2 are the like terms.
2pq, – 3pq, and – pq are the like terms.
q2, 5q2, and 7q2 are the like terms.
{1 + 2 + 4}[p2] = 7p2
{2 – 3 – 1}[pq] = -2pq
{1 + 5 + 7}[q2] = 13q2

Therefore, A + B + C = 7p2 – 2pq + 13q2

Question 3. Subtract arranging the like terms:
(i) 3a – 4b from 7a – 9b + 3
(ii) 4a3 + 2a2b + 3c3 from 10a3 – 3a2b – 5c3
(iii) 6x – 3y – z from 8x – 7y – 3z
(iv) From 5l3 – 2m3 + 4n3 subtract 7l3 + m3 – n3
(v) From 9x2 – xy + 3y3 subtract 3x2 – 3xy – 2y3
(vi) From 8a4 + 5ab + 4b3 subtract 6a4 – 3ab – 2b3

Solution:

(i) Given that subtract 3a – 4b from 7a – 9b + 3.
(7a – 9b + 3) – (3a – 4b) = 7a – 9b + 3 – 3a + 4b
Find out the like terms from the above equation.
7a and 3a are like terms.
9b and 4b are the like terms.
Now, do the subtraction.
7a – 3a – 9b + 4b + 3 = 4a – 5b + 3.

Therefore, the answer is 4a – 5b + 3.

(ii) Given that subtract 4a3 + 2a2b + 3c3 from 10a3 – 3a2b – 5c3.
(10a3 – 3a2b – 5c3) – (4a3 + 2a2b + 3c3) = 10a3 – 3a2b – 5c3 – 4a3 – 2a2b – 3c3
Find out the like terms from the above equation.
10a3 and 4a3 are like terms.
3a2b and 2a2b are the like terms.
5c3 and 3c3 are the like terms.
Now, do the subtraction.
10a3 – 4a3 – 3a2b – 2a2b – 5c3 – 3c3 = 6a3 – 5a2b – 8c3.

Therefore, the answer is 6a3 – 5a2b – 8c3.

(iii) Given that subtract 6x – 3y – z from 8x – 7y – 3z
(8x – 7y – 3z) – (6x – 3y – z) = 8x – 7y – 3z – 6x + 3y + z
Find out the like terms from the above equation.
8x and 6x are like terms.
7y and 3y are the like terms.
z and 3z are the like terms.
Now, do the subtraction.
8x – 6x – 7y + 3y – 3z + z = 2x – 4y – 2z.

Therefore, the answer is 2x – 4y – 2z.

(iv) Given that From 5l3 – 2m3 + 4n3 subtract 7l3 + m3 – n3.
(5l3 – 2m3 + 4n3) – (7l3 + m3 – n3) = 5l3 – 2m3 + 4n3 – 7l3 – m3 + n3
Find out the like terms from the above equation.
7l3 and 5l3 are like terms.
m3 and 2m3 are the like terms.
n3 and 4n3 are the like terms.
Now, do the subtraction.
5l3 – 7l3 – 2m3 – m3 + 4n3 + n3 = -2l3 – 3m3 + 5n3.

Therefore, the answer is -2l3 – 3m3 + 5n3.

(v) Given that From 9x2 – xy + 3y3 subtract 3x2 – 3xy – 2y3.
(9x2 – xy + 3y3) – (3x2 – 3xy – 2y3) = 9x2 – xy + 3y3 – 3x2 + 3xy + 2y3
Find out the like terms from the above equation.
3x2 and 9x2 are like terms.
3xy and xy are the like terms.
2y3 and 3y3 are the like terms.
Now, do the subtraction.
9x2 – 3x2 – xy + 3xy + 3y3 + 2y3 = 6x2 + 2xy + 5y3.

Therefore, the answer is 6x2 + 2xy + 5y3.

(vi) Given that From 8a4 + 5ab + 4b3 subtract 6a4 – 3ab – 2b3.
(8a4 + 5ab + 4b3) – (6a4 – 3ab – 2b3) = 8a4 + 5ab + 4b3 – 6a4 + 3ab + 2b3
Find out the like terms from the above equation.
8a4 and 6a4 are like terms.
5ab and 3ab are the like terms.
4b3 and 2b3 are the like terms.
Now, do the subtraction.
8a4 – 6a4 + 5ab + 3ab + 4b3 + 2b3 = 2a4 + 8ab + 6b3.

Therefore, the answer is 2a4 + 8ab + 6b3

Question 4. The sum of two expressions is l – 2m + 3n if one of the terms is 5l – 4m – n, find the other.

Solution:

Given that the sum of two expressions is l – 2m + 3n and one of the terms is 5l – 4m – n.
Let the second term is x. Then,
(l – 2m + 3n) – (5l – 4m – n) = x
l – 2m + 3n – 5l + 4m + n = x
-4l + 2m + 3n = x

Therefore, the second term is -4l + 2m + 3n.

Question 5. What should be subtracted from -12x + 4y – 2z to obtain 11x – 18y + 5z.

Solution:

Let the other expression be a.
If a is subtracted from -12x + 4y – 2z, we need to get 11x – 18y + 5z.
(-12x + 4y – 2z) – (a) = 11x – 18y + 5z
(-12x + 4y – 2z) – (11x – 18y + 5z) = a
Find out the like terms from the above equation.
12x and 11x are like terms.
4y and 18y are like terms.
2z and 5z are like terms
-12x – 11x + 4y + 18y – 2z – 5z = a
-23x + 22y  – 7z = a

Therefore, -23x + 22y  – 7z needs to subtract from -12x + 4y – 2z to obtain 11x – 18y + 5z.

Question 6. How much is 2a3 + 3b3 + 4c3 less than -8a3 + 4b3 – 5c3?

Solution:

Subtract 2a3 + 3b3 + 4c3 from -8a3 + 4b3 – 5c3.
(-8a3 + 4b3 – 5c3) – (2a3 + 3b3 + 4c3) = -8a3 + 4b3 – 5c3 – 2a3 – 3b3 – 4c3
Find out the like terms from the above equation.
8a3 and 2a3 are like terms.
4b3 and 3b3 are like terms.
5c3 and 4c3 are like terms.
-8a3 – 2a3 + 4b3  – 3b3 – 5c3 – 4c3 = -10a3 + b3 – 9c3

Therefore, the answer is -10a3 + b3 – 9c3.

Question 7. To the sum of 4p – 8q – 3r and 5p – q + 9r at the sum of p – 4q + 7r and r – 6q.

Solution:

Given that to the sum of 4p – 8q – 3r and 5p – q + 9r at the sum of p – 4q + 7r and r – 6q.
Firstly, find the sum of the 4p – 8q – 3r and 5p – q + 9r.
(4p – 8q – 3r) + (5p – q + 9r) = 4p – 8q – 3r + 5p – q + 9r
Find out the like terms from the above equation.
4p and 5p are like terms.
8q and q are like terms.
3r and 9r are like terms.
4p + 5p – 8q – q – 3r + 9r = 9p – 9q + 6r

Therefore, the answer is 9p – 9q + 6r.

Question 8. Subtract 7m – 9m3 + 2m2 from the sum of 4 + 10m2 – 2m3 and 4m3 – 5m2 + 2m – 4.

Solution:

Given that Subtract 7m – 9m3 + 2m2 from the sum of 4 + 10m2 – 2m3 and 4m3 – 5m2 + 2m – 4.
Firstly, find the sum of the 4 + 10m2 – 2m3 and 4m3 – 5m2 + 2m – 4.
(4 + 10m2 – 2m3) + (4m3 – 5m2 + 2m – 4) = 4 + 10m2 – 2m3 + 4m3 – 5m2 + 2m – 4
Find out the like terms from the above equation.
2m3 and 4m3 are like terms.
10m2 and 5m2 are like terms.
4 and -4 are like terms.
– 2m3 + 4m3 – 5m2 + 10m2 + 2m + 4 – 4 = 2m3 + 5m2 + 2m.
Subtract 7m – 9m3 + 2m2 from 2m3 + 5m2 + 2m
(2m3 + 5m2 + 2m) – (7m – 9m3 + 2m2) = 2m3 + 5m2 + 2m – 7m + 9m3 – 2m2
Find out the like terms from the above equation.
2m3 and 9m3 are like terms.
5m2 and 2m2 are like terms.
2m and 7m are like terms.
Now, do the subtraction.
2m3 + 9m3 + 5m2 – 2m2 + 2m – 7m = 11m3 + 3m2 – 5m

Therefore, the answer is 11m3 + 3m2 – 5m.

Question 9. What should be added to 5a3 – 3a2b + 2ab2 – 7b3 to get 9a3 + 5a2b – 3ab2 – 2b3?

Solution:

Let the other expression is x.
By adding x to 5a3 – 3a2b + 2ab2 – 7b3, we will get 9a3 + 5a2b – 3ab2 – 2b3.
x + (5a3 – 3a2b + 2ab2 – 7b3) = 9a3 + 5a2b – 3ab2 – 2b3.
x = (9a3 + 5a2b – 3ab2 – 2b3) – (5a3 – 3a2b + 2ab2 – 7b3)
Now, x = 9a3 + 5a2b – 3ab2 – 2b3 – 5a3 + 3a2b – 2ab2 + 7b3
Find out the like terms from the above equation.
9a3 and 5a3 are like terms.
5a2b and 3a2b are like terms.
3ab2 and 2ab2 are like terms.
2b3 and 7b3 are like terms.
Now, do the subtraction.
x = 9a3 – 5a3 + 5a2b + 3a2b – 3ab2 – 2ab2 – 2b3 + 7b3
x = 4a3 + 8a2b – 5ab2 + 5b3

Therefore, the answer is 4a3 + 8a2b – 5ab2 + 5b3.

Question 10. What should be added to 4p3 – 6p2 + 10p – 12 to get 0?

Solution:

Given that.4p3 – 6p2 + 10p – 12
Let the other expression is a.
To get a + (4p3 – 6p2 + 10p – 12) = 0.
a = 0 – (4p3 – 6p2 + 10p – 12)
a = -4p3 + 6p2 – 10p + 12

Therefore, the answer is -4p3 + 6p2 – 10p + 12.

Question 11. What should be subtracted from 9m3 – 2m2n – 4m2 + 7 to get 0?

Solution:

Given that.9m3 – 2m2n – 4m2 + 7
Let the other expression is a.
To get (9m3 – 2m2n – 4m2 + 7) – (a) = 0.
a = (9m3 – 2m2n – 4m2 + 7) – 0
a = 9m3 – 2m2n – 4m2 + 7

Therefore, the answer is 9m3 – 2m2n – 4m2 + 7.

Question 12. Evaluate the following unlike terms:
(i) 12m – 2n – 10m
(ii) 6pq – 5pq + 3qr
(iii) 3a – 4a + 7b + 12a
(iv) 6m + 2n – 3m – 4n
(v) 9ab + 4ba + 26
(vi) 5xyz – 13xy + 24xyz

Solution:

(i) Given that 12m – 2n – 10m
From the given equation 12m and 10m are the like terms.
12m – 10m – 2n = 2m – 2n.

Therefore, the answer is 2m – 2n.

(ii) Given that 6pq – 5pq + 3qr
From the given equation 6pq and 5pq are the like terms.
6pq – 5pq + 3qr = pq + 3qr

Therefore, the answer is pq + 3qr.

(iii) Given that 3a – 4a + 7b + 12a
From the given equation 3a, 4a, and 12a are the like terms.
3a – 4a + 7b + 12a = 3a – 4a + 12a + 7b = 11a + 7b

Therefore, the answer is 11a + 7b.

(iv) Given that 6m + 2n – 3m – 4n
From the given equation 6m and 3m are the like terms.
2n and 4n are the like terms.
6m + 2n – 3m – 4n = 6m – 3m + 2n – 4n = 3m – 2n

Therefore, the answer is 3m – 2n.

(v) Given that 9ab + 4ba + 26
4ba can also write as 4ab.
From the given equation 9ab and 4ab are the like terms.
9ab + 4ba + 26 = 13ab + 26

Therefore, the answer is 13ab + 26.

(vi) Given that 5xyz – 13xy + 24xyz
From the given equation 5xyz and 24xyz are the like terms.
5xyz – 13xy + 24xyz = 5xyz + 24xyz – 13xy = 29xyz – 13xy

Therefore, the answer is 29xyz – 13xy.

Question 13. What is the difference between -13mn and 7m?

Solution:

Given that -13mn and 7m where m and n are variables.
-13mn and 7m are unlike terms.

Therefore, the answer is -13mn – 7m.

Question 14. How much is 6a + 3b greater than b?

Solution:

To find how much 6a + 3b is greater than b, we need to subtract b from 6a + 3b.
6a + 3b – b = 3a + 2b.

Therefore, the answer is 3a + 2b.

Question 15. What is the sum of -15ab and -12a?

Solution:

Given that -15ab and -12a
From the given equation 15ab and 12a are unlike terms.
-15ab – 12a

Therefore, the sum of -15ab and -12a is -15ab – 12a.

Question 16. I. Subtract the following unlike terms
(i) from 3mn take m
(ii) from 15a2bc take 2abc
(iii) from -4bc take 8ac
II. Subtract the first term from the second term:
(i) 8a, 10b
(ii) 2ab, 4b2
(iii) 6y, 7x

Solution:

(i) Given that subtract m from 3mn.
From the given equation m and 3mn are unlike terms.
3mn – m

Therefore, the answer is 3mn – m.

(ii) Given that subtract 2abc from 15a2bc.
From the given equation 2abc and 15a2bc are unlike terms.
15a2bc – 2abc

Therefore, the answer is 15a2bc – 2abc

(iii) Given that subtract 8ac from -4bc.
From the given equation 8ac and -4bc are unlike terms.
-4bc – 8ac

Therefore, the answer is -4bc – 8ac

II. (i) Given that subtract 8a from 10b.
From the given equation 8a and 10b are unlike terms.
10b – 8a

Therefore, the answer is 10b – 8a

(ii) Given that subtract 2ab from 4b2.
From the given equation 4b2 and 2ab are unlike terms.
4b2 – 2ab

Therefore, the answer is 4b2 – 2ab

(iii) Given that subtract 6y from 7x.
From the given equation 6y and 7x are unlike terms.
7x – 6y

Therefore, the answer is 7x – 6y.

Question 17. John has m number of chocolates and Sam has n number of chocolates. How many chocolates do both of them have altogether?

Solution:

Given that John has m number of chocolates and Sam has n number of chocolates.
Add the number of chocolates John has and the number of chocolates Sam has to find the chocolates they have altogether.
m number of chocolates + n number of chocolates = m + n

Therefore, John and Sam have m + n chocolates together.

Question 18. By how much does 15m3 – 3m2 + m – 1 exceeded 12m3 – m2 + m + 1.

Solution:

Given that 15m3 – 3m2 + m – 1 exceeded 12m3 – m2 + m + 1.
To know how much 15m3 – 3m2 + m – 1 exceeded 12m3 – m2 + m + 1, we need to subtract them.
(15m3 – 3m2 + m – 1) – (12m3 – m2 + m + 1) = 15m3 – 3m2 + m – 1 – 12m3 + m2 – m – 1
Find out the like terms from the above equation.
15m3 and 12m3 are like terms.
3m2 and m2 are like terms.
m and m are like terms.
Now, do the subtraction.
15m3 – 12m3 – 3m2 + m2 + m – m – 1 – 1 = 3m3 – 2m2 – 2.

Therefore, the answer is 3m3 – 2m2 – 2.

Question 19. If the perimeter of a triangle is 8a + 34b + 12c and the two sides are 8a – 6b + 10c, 6a + 14b – 4c, find the third side.

Solution:

Given that the perimeter of a triangle is 8a + 34b + 12c and the two sides are 8a – 6b + 10c, 6a + 14b – 4c.
The perimeter of a triangle is the sum of the sides of the triangle.
let the third side of the triangle is x.
8a + 34b + 12c = 8a – 6b + 10c + 6a + 14b – 4c + x.
8a + 34b + 12c = 14a + 8b + 6c + x
x = (8a + 34b + 12c) – (14a + 8b + 6c)
x = 8a + 34b + 12c – 14a – 8b – 6c
Find out the like terms from the above equation.
8a and 14a are like terms.
34b and 8b are like terms.
12c and 6c are like terms.
x = 8a – 14a + 34b – 8b + 12c – 6c = -6a + 26b + 6c

Therefore, the third side of the triangle is -6a + 26b + 6c.

Question 20. The two sides of a rectangle are 4p3 – 6q3 + 4r3 and -10p3 + 4q3 – 4r3. Find its perimeter.

Solution:

Given that the two sides of a rectangle are 4p3 – 6q3 + 4r3 and -10p3 + 4q3 – 4r3.
The perimeter of a rectangle is the sum of the sides of the triangle. A rectangle consists of opposite sides that are parallel and equal to each other.
The perimeter of a rectangle = 2 (4p3 – 6q3 + 4r3) + 2 (-10p3 + 4q3 – 4r3.) = 8p3 – 12q3 + 8r3 -20p3 + 8q3 – 8r3
Find out the like terms from the above equation.
8p3 and 20p3 are like terms.
12q3 and 8q3 are like terms.
8r3 and 8r3 are like terms.
The perimeter of a rectangle = 8p3 -20p3 – 12q3 + 8q3 + 8r3 – 8r3 = -12p3 – 4q3

Therefore, the perimeter of a rectangle is -12p3 – 4q3.

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