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## Free Like and Unlike Terms Worksheet with Answers

**Question 1.**

Add arranging the like terms

(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 = p^{2} + 2pq + q^{2}, B = 2p^{2} â€“ 3pq + 5q^{2} and C = 4p^{2} â€“ pq – 7q^{2}, find A + B + C.

**Solution:**

Given that A = p^{2} + 2pq + q^{2}, B = 2p^{2} â€“ 3pq + 5q^{2} and C = 4p^{2} â€“ pq – 7q^{2}.

Now, add A + B + C = (p^{2} + 2pq + q^{2}) + (2p^{2} â€“ 3pq + 5q^{2}) + (4p^{2} â€“ pq – 7q^{2})

From the above equation, p^{2}, 2p^{2}, and 4p^{2} are the like terms.

2pq, â€“ 3pq, and â€“ pq are the like terms.

q^{2}, 5q^{2}, and 7q^{2}Â are the like terms.

Add the coefficients of p^{2}.

{1 + 2 + 4}[p^{2}] = 7p^{2}

Add the coefficients of pq.

{2 – 3 – 1}[pq] = -2pq^{
}Add the coefficients of q^{2}.

{1 + 5 + 7}[q^{2}] = 13q^{2}

Therefore, A + B + C = 7p^{2} – 2pq + 13q^{2}

**Question 3.** Subtract arranging the like terms:

(i) 3a â€“ 4b from 7a â€“ 9b + 3

(ii) 4a^{3} + 2a^{2}b + 3c^{3} from 10a^{3} â€“ 3a^{2}b â€“ 5c^{3}

(iii) 6x â€“ 3y – z from 8x â€“ 7y â€“ 3z

(iv) From 5l^{3} â€“ 2m^{3} + 4n^{3} subtract 7l^{3} + m^{3} â€“ n^{3}

(v) From 9x^{2} â€“ xy + 3y^{3} subtract 3x^{2} â€“ 3xy â€“ 2y^{3}

(vi) From 8a^{4} + 5ab + 4b^{3} subtract 6a^{4} â€“ 3ab â€“ 2b^{3}

**Solution:**

(i) Given that subtract 3a â€“ 4b from 7a â€“ 9b + 3.

(7a â€“ 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 4a^{3} + 2a^{2}b + 3c^{3} from 10a^{3} â€“ 3a^{2}b â€“ 5c^{3}.

(10a^{3} â€“ 3a^{2}b â€“ 5c^{3}) – (4a^{3} + 2a^{2}b + 3c^{3}) = 10a^{3} â€“ 3a^{2}b â€“ 5c^{3} – 4a^{3} – 2a^{2}b – 3c^{3}

Find out the like terms from the above equation.

10a^{3} and 4a^{3} are like terms.

3a^{2}b and 2a^{2}b are the like terms.

5c^{3} and 3c^{3}Â are the like terms.

Now, do the subtraction.

10a^{3} – 4a^{3} â€“ 3a^{2}b – 2a^{2}b â€“ 5c^{3} – 3c^{3} = 6a^{3} – 5a^{2}b – 8c^{3}.

Therefore, the answer is 6a^{3} – 5a^{2}b – 8c^{3}.

(iii) Given that subtract 6x â€“ 3y – z from 8x â€“ 7y â€“ 3z

(8x â€“ 7y â€“ 3z) – (6x â€“ 3y – z) = 8x – 7y – 3z – 6x + 3y + z

Find out the like terms from the above equation.

8x and 6xÂ are like terms.

7y and 3yÂ are the like terms.

z and 3zÂ 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 5l^{3} â€“ 2m^{3} + 4n^{3} subtract 7l^{3} + m^{3} â€“ n^{3}.

(5l^{3} â€“ 2m^{3} + 4n^{3}) – (7l^{3} + m^{3} â€“ n^{3}) = 5l^{3} â€“ 2m^{3} + 4n^{3 }– 7l^{3} – m^{3} + n^{3}

Find out the like terms from the above equation.

7l^{3} and 5l^{3} are like terms.

m^{3} and 2m^{3}Â are the like terms.

n^{3} and 4n^{3}Â are the like terms.

Now, do the subtraction.

5l^{3} – 7l^{3 }â€“ 2m^{3} – m^{3} + 4n^{3 }+ n^{3} = -2l^{3} – 3m^{3} + 5n^{3}.

Therefore, the answer is -2l^{3} – 3m^{3} + 5n^{3}.

(v) Given that From 9x^{2} â€“ xy + 3y^{3} subtract 3x^{2} â€“ 3xy â€“ 2y^{3}.

(9x^{2} â€“ xy + 3y^{3}) – (3x^{2} â€“ 3xy â€“ 2y^{3}) = 9x^{2} â€“ xy + 3y^{3 }– 3x^{2} + 3xy + 2y^{3}

Find out the like terms from the above equation.

3x^{2} and 9x^{2}Â are like terms.

3xy and xyÂ are the like terms.

2y^{3} and 3y^{3} are the like terms.

Now, do the subtraction.

9x^{2} – 3x^{2 }â€“ xy + 3xy + 3y^{3} + 2y^{3} = 6x^{2} + 2xy + 5y^{3}.

Therefore, the answer is 6x^{2} + 2xy + 5y^{3}.

(vi) Given that From 8a^{4} + 5ab + 4b^{3} subtract 6a^{4} â€“ 3ab â€“ 2b^{3}.

(8a^{4} + 5ab + 4b^{3}) – (6a^{4} â€“ 3ab â€“ 2b^{3}) = 8a^{4} + 5ab + 4b^{3} – 6a^{4} + 3ab + 2b^{3}

Find out the like terms from the above equation.

8a^{4} and 6a^{4} are like terms.

5ab and 3ab are the like terms.

4b^{3} and 2b^{3}Â are the like terms.

Now, do the subtraction.

8a^{4} – 6a^{4} + 5ab + 3ab + 4b^{3} + 2b^{3} = 2a^{4} + 8ab + 6b^{3}.

Therefore, the answer is 2a^{4} + 8ab + 6b^{3}

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

**Solution:**

Given that the sum of two expressions is l â€“ 2m + 3n and one of the terms is 5l â€“ 4m â€“ n.

Let the second term is x. Then,

(l â€“ 2m + 3n) – (5l â€“ 4m â€“ n) = x

l â€“ 2m + 3n – 5l + 4m + n = x

-4l + 2m + 3n = x

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

**Question 5.Â **What should be subtracted from -12x + 4y â€“ 2z to obtain 11x â€“ 18y + 5z.

**Solution:**

Let the other expression be a.

If a is subtracted from -12x + 4y â€“ 2z, we need to get 11x â€“ 18y + 5z.

(-12x + 4y â€“ 2z) – (a) = 11x â€“ 18y + 5z

(-12x + 4y â€“ 2z) – (11x â€“ 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 + 22yÂ – 7z = a

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

Do Read: Worksheet on Combining Like Terms

**Question 6. **How much is 2a^{3} + 3b^{3} + 4c^{3} less than -8a^{3} + 4b^{3} â€“ 5c^{3}?

**Solution:**

Subtract 2a^{3} + 3b^{3} + 4c^{3} from -8a^{3} + 4b^{3} â€“ 5c^{3}.

(-8a^{3} + 4b^{3} â€“ 5c^{3}) – (2a^{3} + 3b^{3} + 4c^{3}) = -8a^{3} + 4b^{3} â€“ 5c^{3} – 2a^{3} – 3b^{3} – 4c^{3}

Find out the like terms from the above equation.

8a^{3 }and 2a^{3 }are like terms.

4b^{3} and 3b^{3} are like terms.

5c^{3} and 4c^{3} are like terms.

-8a^{3} – 2a^{3 }+ 4b^{3} Â – 3b^{3 }â€“ 5c^{3} – 4c^{3} = -10a^{3} + b^{3} â€“ 9c^{3}

Therefore, the answer is -10a^{3} + b^{3} â€“ 9c^{3}.

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

**Solution:**

Given that to the sum of 4p â€“ 8q – 3r and 5p â€“ q + 9r at the sum of p â€“ 4q + 7r and r â€“ 6q.

Firstly, find the sum of the 4p â€“ 8q – 3r and 5p â€“ q + 9r.

(4p â€“ 8q – 3r) + (5p â€“ q + 9r) = 4p â€“ 8q – 3r + 5p â€“ 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 â€“ 8q â€“ q – 3r + 9r = 9p – 9q + 6r

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

**Question 8. **Subtract 7m â€“ 9m^{3} + 2m^{2} from the sum of 4 + 10m^{2} â€“ 2m^{3} and 4m^{3} â€“ 5m^{2} + 2m â€“ 4.

**Solution:**

Given that Subtract 7m â€“ 9m^{3} + 2m^{2} from the sum of 4 + 10m^{2} â€“ 2m^{3} and 4m^{3} â€“ 5m^{2} + 2m â€“ 4.

Firstly, find the sum of the 4 + 10m^{2} â€“ 2m^{3} and 4m^{3} â€“ 5m^{2} + 2m â€“ 4.

(4 + 10m^{2} â€“ 2m^{3}) + (4m^{3} â€“ 5m^{2} + 2m â€“ 4) = 4 + 10m^{2} â€“ 2m^{3} + 4m^{3} â€“ 5m^{2} + 2m â€“ 4

Find out the like terms from the above equation.

2m^{3} and 4m^{3} are like terms.

10m^{2} and 5m^{2} are like terms.

4 and -4 are like terms.

Now, do the addition.

â€“ 2m^{3} + 4m^{3} â€“ 5m^{2} + 10m^{2} + 2m + 4 â€“ 4 = 2m^{3} + 5m^{2} + 2m.

Subtract 7m â€“ 9m^{3} + 2m^{2} from 2m^{3} + 5m^{2} + 2m

(2m^{3} + 5m^{2} + 2m) – (7m â€“ 9m^{3} + 2m^{2}) = 2m^{3} + 5m^{2} + 2m – 7m + 9m^{3} – 2m^{2}

Find out the like terms from the above equation.

2m^{3} and 9m^{3} are like terms.

5m^{2} and 2m^{2 }are like terms.

2m and 7m are like terms.

Now, do the subtraction.

2m^{3} + 9m^{3 }+ 5m^{2} – 2m^{2} + 2m – 7m = 11m^{3} + 3m^{2} – 5m

Therefore, the answer is 11m^{3} + 3m^{2} – 5m.

**Question 9.Â **What should be added to 5a^{3} â€“ 3a^{2}b + 2ab^{2} â€“ 7b^{3} to get 9a^{3} + 5a^{2}b â€“ 3ab^{2} â€“ 2b^{3}?

**Solution:**

Let the other expression is x.

By adding x to 5a^{3} â€“ 3a^{2}b + 2ab^{2} â€“ 7b^{3}, we will get 9a^{3} + 5a^{2}b â€“ 3ab^{2} â€“ 2b^{3}.

x + (5a^{3} â€“ 3a^{2}b + 2ab^{2} â€“ 7b^{3}) = 9a^{3} + 5a^{2}b â€“ 3ab^{2} â€“ 2b^{3}.

x = (9a^{3} + 5a^{2}b â€“ 3ab^{2} â€“ 2b^{3}) – (5a^{3} â€“ 3a^{2}b + 2ab^{2} â€“ 7b^{3})

Now, x = 9a^{3} + 5a^{2}b â€“ 3ab^{2} â€“ 2b^{3} – 5a^{3} + 3a^{2}b – 2ab^{2} + 7b^{3}

Find out the like terms from the above equation.

9a^{3} and 5a^{3} are like terms.

5a^{2}b and 3a^{2}b are like terms.

3ab^{2} and 2ab^{2} are like terms.

2b^{3 }and 7b^{3 }are like terms.

Now, do the subtraction.

x = 9a^{3} – 5a^{3} + 5a^{2}b + 3a^{2}b â€“ 3ab^{2} – 2ab^{2} â€“ 2b^{3} + 7b^{3}

x = 4a^{3} + 8a^{2}b â€“ 5ab^{2} + 5b^{3}

Therefore, the answer is 4a^{3} + 8a^{2}b â€“ 5ab^{2} + 5b^{3}.

**Question 10. **What should be added to 4p^{3} â€“ 6p^{2} + 10p â€“ 12 to get 0?

**Solution:**

Given that.4p^{3} â€“ 6p^{2} + 10p â€“ 12

Let the other expression is a.

To get a + (4p^{3} â€“ 6p^{2} + 10p â€“ 12) = 0.

a = 0 – (4p^{3} â€“ 6p^{2} + 10p â€“ 12)

a = -4p^{3} + 6p^{2} – 10p + 12

Therefore, the answer is -4p^{3} + 6p^{2} – 10p + 12.

**Question 11.Â **What should be subtracted from 9m^{3} â€“ 2m^{2}n â€“ 4m^{2} + 7 to get 0?

**Solution:**

Given that.9m^{3} â€“ 2m^{2}n â€“ 4m^{2} + 7

Let the other expression is a.

To get (9m^{3} â€“ 2m^{2}n â€“ 4m^{2} + 7) – (a) = 0.

a = (9m^{3} â€“ 2m^{2}n â€“ 4m^{2} + 7) – 0

a = 9m^{3} â€“ 2m^{2}n â€“ 4m^{2} + 7

Therefore, the answer is 9m^{3} â€“ 2m^{2}n â€“ 4m^{2} + 7.

**Question 12.Â **Evaluate the following unlike terms:

(i) 12m â€“ 2n â€“ 10m

(ii) 6pq â€“ 5pq + 3qr

(iii) 3a â€“ 4a + 7b + 12a

(iv) 6m + 2n â€“ 3m â€“ 4n

(v) 9ab + 4ba + 26

(vi) 5xyz â€“ 13xy + 24xyz

**Solution:**

(i) Given that 12m â€“ 2n â€“ 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 â€“ 5pq + 3qr

From the given equation 6pq and 5pq are the like terms.

6pq â€“ 5pq + 3qr = pq + 3qr

Therefore, the answer is pq + 3qr.

(iii) Given that 3a â€“ 4a + 7b + 12a

From the given equation 3a, 4a, and 12a are the like terms.

3a â€“ 4a + 7b + 12a = 3a – 4a + 12a + 7b = 11a + 7b

Therefore, the answer is 11a + 7b.

(iv) Given that 6m + 2n â€“ 3m â€“ 4n

From the given equation 6m and 3m are the like terms.

2n and 4n are the like terms.

6m + 2n â€“ 3m â€“ 4n = 6m â€“ 3m + 2n â€“ 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 â€“ 13xy + 24xyz

From the given equation 5xyz and 24xyz are the like terms.

5xyz â€“ 13xy + 24xyz = 5xyz + 24xyz – 13xy = 29xyz – 13xy

Therefore, the answer is 29xyz – 13xy.

**Question 13.Â **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.Â **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.Â **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 15a^{2}bc take 2abc

(iii) from -4bc take 8ac

II. Subtract the first term from the second term:

(i) 8a, 10b

(ii) 2ab, 4b^{2
}(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 15a^{2}bc.

From the given equation 2abc and 15a^{2}bc are unlike terms.

15a^{2}bc – 2abc

Therefore, the answer is 15a^{2}bc – 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 4b^{2}.

From the given equation 4b^{2} and 2ab are unlike terms.

4b^{2} – 2ab

Therefore, the answer is 4b^{2} – 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.Â **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 15m^{3} â€“ 3m^{2} + m â€“ 1 exceeded 12m^{3} â€“ m^{2} + m + 1.

**Solution:**

Given that 15m^{3} â€“ 3m^{2} + m â€“ 1 exceeded 12m^{3} â€“ m^{2} + m + 1.

To know how much 15m^{3} â€“ 3m^{2} + m â€“ 1 exceeded 12m^{3} â€“ m^{2} + m + 1, we need to subtract them.

(15m^{3} â€“ 3m^{2} + m â€“ 1) – (12m^{3} â€“ m^{2} + m + 1) = 15m^{3} â€“ 3m^{2} + m â€“ 1 – 12m^{3} + m^{2} – m – 1

Find out the like terms from the above equation.

15m^{3} and 12m^{3}Â are like terms.

3m^{2} and m^{2} are like terms.

m and mÂ are like terms.

Now, do the subtraction.

15m^{3} – 12m^{3 }â€“ 3m^{2} + m^{2} + m – m â€“ 1 – 1 = 3m^{3} â€“ 2m^{2} â€“ 2.

Therefore, the answer is 3m^{3} â€“ 2m^{2} â€“ 2.

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

**Solution:**

Given that the perimeter of a triangle is 8a + 34b + 12c and the two sides are 8a â€“ 6b + 10c, 6a + 14b â€“ 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 â€“ 6b + 10c + 6a + 14b â€“ 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 4p^{3} â€“ 6q^{3} + 4r^{3} and -10p^{3} + 4q^{3} â€“ 4r^{3}. Find its perimeter.

**Solution:**

Given that the two sides of a rectangle are 4p^{3} â€“ 6q^{3} + 4r^{3} and -10p^{3} + 4q^{3} â€“ 4r^{3}.

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 (4p^{3} â€“ 6q^{3} + 4r^{3}) + 2 (-10p^{3} + 4q^{3} â€“ 4r^{3}.) = 8p^{3} â€“ 12q^{3} + 8r^{3} -20p^{3} + 8q^{3} â€“ 8r^{3}

Find out the like terms from the above equation.

8p^{3} and 20p^{3} are like terms.

12q^{3} and 8q^{3} are like terms.

8r^{3} and 8r^{3 }are like terms.

The perimeter of a rectangle = 8p^{3} -20p^{3 }â€“ 12q^{3} + 8q^{3} + 8r^{3} â€“ 8r^{3} = -12p^{3} â€“ 4q^{3}

Therefore, the perimeter of a rectangle is -12p^{3} â€“ 4q^{3}.