SUBTRACT THE ALGEBRAIC EXPRESSION PRACTICE

How to subtract the algebraic expression ?

Step 1 :

Write all the expressions into brackets and put the subtraction sign in between.

Step 2 :

Combine all the like terms together from all the expressions and rewrite them in a single expression.

Step 3 :

Add or subtract numerical coefficients of all the like terms followed by the common variable.

Step 4 :

Rewrite the simplified expression, and make sure all the terms in the final answer are unlike terms.

Subtract :

Problem 1 :

14x2 – 12x – 6 from the sum of 2x2 – 3x + 11 and 7x2 – 3x – 4.

Solution :

Let,

a  =  2x2 – 3x + 11

b  =  7x2 – 3x – 4

c  =  14x2 – 12x – 6

The sum of a + b  =  (2x2 – 3x + 11) + (7x2 – 3x – 4)

By combining like terms ,

=  (2x2 + 7x2) + (-3x – 3x) + (11 – 4)

a + b =  9x2 - 6x + 7

(a + b) – c  =  (9x2 - 6x + 7) – (14x2 – 12x – 6)

=  9x2 – 6x + 7 – 14x2 + 12x + 6

By combining like terms ,

 =  (9x2 – 14x2) + (-6x + 12x) + (7 + 6)

=  -5x2 + 6x + 13

So, the answer is  -5x2 + 6x + 13.

Problem 2 :

2x3 + 7x2 – 4x – 13 from the sum of 17x3 – 12x + 11 and x3 + 5.

Solution :

Let,

a  =  17x3 – 12x + 11

b  =  x3 + 5

c  =  2x3 + 7x2 – 4x – 13   

The sum of a + b  =  (17x3 – 12x + 11) + (x3 + 5)

By combining like terms,

=  (17x3 + x3) + (-12x) + (11 + 5)

a + b =  18x3 – 12x + 16

(a + b) – c  =  (18x3 – 12x + 16) – (2x3 + 7x2 – 4x – 13)   

=  18x3 – 12x + 16 – 2x3 – 7x2 + 4x + 13

By combining like terms,

=  (18x3 – 2x3) – (7x2) + (-12x + 4x) + (16 + 13)

=  16x3 – 7x2 – 8x + 29

So, the answer is 16x3 – 7x2 – 8x + 29.

Problem 3 :

The sum of w2 – 10w + 25 and 3w2 – 7w from the product of 4 and 2 – 5w.

Solution :

Let,  

a  =  w2 – 10w + 25

b  =  3w2 – 7w

c  =  4

d  =  2 - 5w

The sum of a + b  =  (w2 – 10w + 25) + (3w2 – 7w)

By combining like terms,

=  (w2 + 3w2) + (-10w – 7w) + 25

a + b  =  4w2 - 17w + 25

The product  of c and d  =  4 × (2 – 5w)

=  8 – 20w

(c × d) – (a + b) =  (8 – 20w) - (4w2 - 17w + 25)

=  8 - 20w - 4w2 + 17w - 25

By combining like terms,

=  (-4w2) + (-20w + 17w) + (8 - 25)

=  -4w2 – 3w - 17    

So, the answer is  -4w2 – 3w – 17.

Problem 4 :

The product of 3 and 6w2 – 7w + 1 from the product of -2 and -3 – 7w.

Solution :

Let,

a  =  3 

b  =  6w2 – 7w + 1  

c  =  -2 

d  =  -3 – 7w

The product of a and b  =  3 x (6w2 – 7w + 1) 

=  18w2 – 21w + 3

The product of c and d  =  -2 x (-3 – 7w)

=  6 + 14w

(c x d) – (a x b)  =  (6 + 14w) – (18w2 – 21w + 3)

=  (6 + 14w) – (18w2 – 21w + 3)

=  6 + 14w – 18w2 + 21w – 3

By combining like terms ,

=  -18w2 + (14w + 21w) + (6 – 3)

=  -18w2 + 35w + 3

So, the answer is -18w2 + 35w + 3.

Simplify :

Problem 5 :

-2 {3a – 4[a – (2 + a)] }

Solution :

Given, -2 {3a – 4[a – (2 + a)] }

=  -2 {3a – 4[a –2 – a] }

=  -2 {3a – 4a + 8 + 4a}

=  -6a + 8a – 16 - 8a

=  -6a - 16 

Problem 6 :

5 {3c – [d – 2(c + d)] }

Solution :

Given, 5 {3c – [d – 2(c + d)] }

=  5 {3c – [d – 2c - 2d] }

=    5 {3c – d + 2c + 2d}

=  15c – 5d + 10c + 10d

=  25c + 5d

Problem 7 :

-3a(2 + b) – 18a(b + 1)

Solution :

Given, -3a(2 + b) – 18a(b + 1)

=  -6a3ab18ab18a

=  -24a – 21ab

Problem 8 :

(-4xy2 – 16xy) [a(3y +2) – 2a(y – 1)]

Solution :

Given,  (-4xy2 – 16xy) [a(3y +2) – 2a(y – 1)]

=  (-4xy2 – 16xy) (3ay + 2a – 2ay + 2a)

=  (-4xy2 – 16xy) (ay + 4a) 

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