RD Sharma Class 7 ex 7.4 Solutions Chapter 7 Algebraic Expressions

In this chapter, we provide RD Sharma Class 7 ex 7.4 Solutions Chapter 7 Algebraic Expressions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 7.4 Solutions Chapter 7 Algebraic Expressions Maths pdf, Now you will get step by step solution to each question.

Maths pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 7
Chapter NameAlgebraic Expressions
ExerciseEx 7.4

RD Sharma Solutions for Class 7 Chapter 7 Decimals Ex 7.4 Download PDF

Chapter 7: Algebraic Expressions Exercise – 7.4

Question: 1

Simplify, the algebraic expressions by removing grouping symbols.

2x + (5x – 3y)

Solution:

We have

2x + (5x – 3y)

Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in the parentheses when we remove them.

= 2x + 5x – 3y

= 7x – 3y

Question: 2

Simplify, the algebraic expressions by removing grouping symbols.

3x – (y – 2x)

Solution:

We have

3x – (y – 2x)

Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

3x – y + 2x

= 5x – y

Question: 3

Simplify, the algebraic expressions by removing grouping symbols.

5a – (3b – 2a + 4c)

Solution:

We have

5a – (3b – 2a + 4c)

Since the ‘-‘ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them.

= 5a – 3b + 2a – 4c

= 7a – 3b – 4c

Question: 4

Simplify, the algebraic expressions by removing grouping symbols.

-2(x2 – y2 + xy) – 3(x2 +y2 – xy)

Solution:

We have

– 2(x2 – y2 + xy) – 3(x2 +y2 – xy)

Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

= -2x+ 2y2 – 2xy – 3x2 – 3y2 + 3xy

= -2x2 – 3x2 + 2y2 – 3y– 2xy + 3xy

= -5x– y2 + xy

Question: 5

Simplify, the algebraic expressions by removing grouping symbols.

3x + 2y – {x – (2y – 3)}

Solution:

We have

3x + 2y – {x – (2y – 3)}

First, we have to remove the small brackets (or parentheses): ( ). Then, we have to remove the curly brackets (or braces): { }.

Therefore,

= 3x + 2y – {x – 2y + 3}

= 3x + 2y – x + 2y – 3

= 2x + 4y – 3

Question: 6

Simplify, the algebraic expressions by removing grouping symbols.

5a – {3a – (2 – a) + 4}

Solution:

We have

5a – {3a – (2 – a) + 4}

First, we have to remove the small brackets (or parentheses): ( ). Then, we have to remove the curly brackets (or braces): { }.

Therefore,

= 5a – {3a – 2 + a + 4}

= 5a – 3a + 2 – a – 4

= 5a – 4a – 2

= a – 2

Question: 7

Simplify, the algebraic expressions by removing grouping symbols.

a – [b – {a – (b – 1) + 3a}]

Solution:

First we have to remove the parentheses, or small brackets, ( ), then the curly brackets, { }, and then the square brackets [ ].

Therefore, we have

a – [b – {a – (b – 1) + 3a}]

= a – [b – {a – b + 1 + 3a}]

= a – [b – {4a – b + 1}]

= a – [b – 4a + b – 1]

= a – [2b – 4a – 1]

= a – 2b + 4a + 1

= 5a – 2b + 1

Question: 8

Simplify, the algebraic expressions by removing grouping symbols.

 a – [2b – {3a – (2b – 3c)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

a – [2b – {3a – (2b – 3c)}]

= a – [2b – {3a – 2b + 3c}]

= a – [2b – 3a + 2b – 3c]

= a – [4b – 3a – 3c]

= a – 4b + 3a + 3c

= 4a – 4b + 3c

Question: 9

Simplify, the algebraic expressions by removing grouping symbols.

 -x + [5y – {2x – (3y – 5x)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets { }, and then the square brackets, [ ].

Therefore, we have

– x + [5y – {2x – (3y – 5x)}]

= – x + [5y – {2x – 3y + 5x)]

= – x + [5y – {7x – 3y}]

= – x + [5y – 7x + 3y]

= – x + [8y – 7x]

= – x + 8y – 7x

= – 8x + 8y

Question: 10

Simplify, the algebraic expressions by removing grouping symbols.

 2a – [4b – {4a – 3(2a – b)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

2a – [4b – {4a – 3(2a – b)}]

= 2a – [4b – {4a – 6a + 3b}]

= 2a – [4b – {- 2a + 3b}]

= 2a – [4b + 2a – 3b]

= 2a – [b + 2a]

= 2a – b – 2a

= – b

Question: 11

Simplify, the algebraic expressions by removing grouping symbols.

-a – [a + {a + b – 2a – (a – 2b)} – b]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

– a – [a + {a + b – 2a – (a – 2b)} – b]

= – a – [a + {a + b – 2a – a + 2b} – b]

= – a – [a + {- 2a + 3b} – b]

= – a – [a – 2a + 3b – b]

= – a – [- a + 2b]

= – a + a – 2b

= – 2b

Question: 12

Simplify, the algebraic expressions by removing grouping symbols.

2x – 3y – [3x – 2y -{x – z – (x – 2y)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

2x – 3y – [3x – 2y – {x – z – (x – 2y)})

= 2x – 3y – [3x – 2y – {x – z – x + 2y}]

= 2x – 3y – [3x – 2y – {- z + 2y}]

= 2x – 3y – [3x – 2y + z – 2y]

= 2x – 3y – [3x – 4y + z]

= 2x – 3y – 3x + 4y – z

= – x + y – z

Question: 13

Simplify, the algebraic expressions by removing grouping symbols.

 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]

= 5 + [x – {2y – 6x – y + 4 + 2x} – {x – y + 2}]

= 5 + [x – {y – 4x + 4} – {x – y + 2}]

= 5 + [x – y + 4x – 4 – x + y – 2]

= 5 + [4x – 6]

= 5 + 4x – 6

= 4x – 1

Question: 14

Simplify, the algebraic expressions by removing grouping symbols.

x– [3x + [2x – (x– 1)] + 2]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

x2 – [3x + [2x – (x2 – 1)] + 2]

= x2 – [3x + [2x – x2 + 1] + 2]

= x2 – [3x + 2x – x2 + 1 + 2]

= x2 – [5x – x2 + 3]

= x2 – 5x + x2 – 3

= 2x2 – 5x – 3

Question: 15

Simplify, the algebraic expressions by removing grouping symbols.

20 – [5xy + 3[x2 – (xy – y) – (x – y)]]

Solution:

20 – [5xy + 3[x2 – (xy – y) – (x – y)]]

= 20 – [5xy + 3[x2 – xy + y – x + y]]

= 20 – [5xy + 3[x2 – xy + 2y – x]]

= 20 – [5xy + 3x2 – 3xy + 6y – 3x]

= 20 – [2xy + 3x2 + 6y – 3x]

= 20 – 2xy – 3x2 – 6y + 3x

= – 3x– 2xy – 6y + 3x + 20

Question: 16

Simplify, the algebraic expressions by removing grouping symbols.

85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]

= 85 – [12x – 56x + 21 – 2{10x – 10 + 20x}]

= 85 – [12x – 56x + 21 – 2{30x – 10}]

= 85 – [12x – 56x + 21 – 60x + 20]

= 85 – [12x – 116x + 41]

= 85 – [- 104x + 41]

= 85 + 104x – 41

= 44 + 104x

Question: 17

Simplify, the algebraic expressions by removing grouping symbols.

 xy[yz – zx – {yx – (3y – xz) – (xy – zy)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

xy – [yz – zx – {yx – (3y – xz) – (xy – zy)}]

= xy – [yz – zx – {yx – 3y + xz – xy + zy}]

= xy – [yz – zx – {- 3y + xz + zy}]

= xy – [yz – zx + 3y – xz – zy]

= xy – [- zx + 3y – xz]

= xy – [- 2zx + 3y]

= xy + 2xz – 3y

All Chapter RD Sharma Solutions For Class 7 Maths

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

Leave a Comment

Your email address will not be published.