PRACTICE PROBLEMS ON SIMPLIFYING ALGEBRAIC EXPRESSIONS

Detailed Solution

Problem 1 :

-4xy + 12xy

Solution :

= -4xy + 12xy

Like terms only are to be  added or subtracted. In the given expression, we have two terms. In both terms, we have same set of variables xy. Then these two terms are to be considered as like terms.

Since these two terms are having different signs, we have to subtract them and put greater number sign for the result. 

= 8xy

Problem 2 :

-7 + x - 5 + 2x

Solution :

= -7 + x - 5 + 2x

Here we have terms with the variable x and constant.

= -7 - 5 + x + 2x

= -12 + 3x

Variable should be written first and constant should be written at last. Then the answer is

= 3x - 12

Problem 3 :

6w² + 11w + 8w² - 15w

Solution :

= 6w² + 11w + 8w² - 15w

Here we see two set of terms. 

• Terms with w²
• terms with w

By writing like terms near by,

= 6w² + 8w² - 15w + 11w

= 14w² - 4w

Problem 4 :

y + y - 10x

Solution :

= y + y - 10x

Here we see two set of terms. 

• Terms with y
• Terms with x

= y + y - 10x

= 2y - 10x

Problem 5 :

-5y - 9y + 5x - 10x + 14y

Solution :

= -5y - 9y + 5x - 10x + 14y

Let us write the terms involving x near by and terms involving y near by.

= -5y - 9y + 14y + 5x - 10x 

= -14y + 14y + 5x - 10x

= 0y - 5x

= -5x

Problem 6 :

6x + 4 + 15 - 7x

Solution :

= 6x + 4 + 15 - 7x

Let us write the terms involving x near by and constants near by.

= 6x - 7x + 4 + 15

= -1x + 19

Problem 7 :

5a + 2b - 5b + 9a

Solution :

= 5a + 2b - 5b + 9a

By observing the expression, two set of terms are there. 

• Terms with variable a
• Terms with variable b

= 5a + 9a + 2b - 5b 

= 14a - 3b

Problem 8 :

j - 4k + 7 + 2j - 3k

Solution :

= j - 4k + 7 + 2j - 3k

By observing the expression, we have three set of terms.

• Terms with variable j
• Terms with variable k
• Constants

Writing like terms near by, we get

= j + 2j - 4k - 3k + 7

= 3j - 7k + 7

Problem 9 :

2c - 3 + 9d + 7 - 6d

Solution :

= 2c - 3 + 9d + 7 - 6d

By observing the expression, we have three set of terms.

• Terms with c
• Terms with d
• Constant

By writing the like terms near by, we get

= 2c - 6d+ 9d - 3 + 7

= 2c + 3d + 4

Problem 10 :

4c + 7 - 3c + 8

Solution :

= 4c + 7 - 3c + 8

We see two set of terms,

• Terms with the variable c
• Constants

= 4c  - 3c + 8 + 7

= c + 15

Problem 11 :

4c + 7 - 3c + 8

Solution :

= 4c + 7 - 3c + 8

Writing the like terms near by, we get

= 4c - 3c + 7 + 8

= c + 15

Problem 12 :

6d - e + 3d - 2e

Solution :

= 6d - e + 3d - 2e

Writing the like terms near by, we get

= 6d + 3d - e - 2e

= 9d - 3e

Problem 13 :

3 + 2c + 5c + 2

Solution :

= 3 + 2c + 5c + 2

Writing the like terms near by, we get

= 3 + 2 + 2c + 5c

= 5 + 7c

So, the answer is 

= 7c + 5

Problem 14 :

2a + 3b - 5b + 7a

Solution :

= 2a + 3b - 5b + 7a

Combining the like terms, we get

= 2a + 7a + 3b - 5b

= 9a - 2b

Problem 15 :

6c - 7 + d + 4 - 6d

Solution :

= 6c - 7 + d + 4 - 6d

Writing the like terms near by,

= 6c + d  - 6d + 4 - 7

= 6c - 5d - 3