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

(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.

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.
Add the coefficients of p².
{1 + 2 + 4}[p²] = 7p²
Add the coefficients of pq.
{2 – 3 – 1}[pq] = -2pq
Add the coefficients of q².
{1 + 5 + 7}[q²] = 13q²

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

(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

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.

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.

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.

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.
Now, do the addition.
4p + 5p – 8q – q – 3r + 9r = 9p – 9q + 6r

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

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.
Now, do the addition.
– 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.

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.

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.

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.

(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.

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

Therefore, the answer is -13mn – 7m.

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.

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.

(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.

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.

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.

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.

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.