The document RD Sharma Solutions (Part -1) - Ex - 7.4, Algebraic Expressions, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**Simplify each of the following algebraic expressions by removing grouping symbols. 2 x + (5x âˆ’ 3y)**

We have

2x + (5x âˆ’ 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

**Simplify each of the following algebraic expressions by removing grouping symbols. 3 x âˆ’ (y âˆ’ 2x)**

We have

3x âˆ’ (y âˆ’ 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 âˆ’ y + 2x

= 5x - y

**Simplify each of the following algebraic expressions by removing grouping symbols. 5 a âˆ’ (3b âˆ’ 2a + 4c)**

We have

5a âˆ’ (3b âˆ’ 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

**Simplify each of the following algebraic expressions by removing grouping symbols. âˆ’2 ( x^{2} âˆ’ y^{2} + xy) âˆ’ 3(x^{2} + y^{2} âˆ’ xy)**

We have

âˆ’ 2(x^{2} âˆ’ y^{2} + xy) âˆ’ 3(x^{2} + y^{2} âˆ’ xy)

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

= - 2x^{2}^{ }+ 2y^{2}^{ }- 2xy - 3x^{2} - 3y^{2}^{ }+ 3xy

= - 2x^{2}^{ }- 3x^{2}^{ }+ 2y^{2}- 3y^{2} - 2xy + 3xy

= - 5x^{2} - y^{2} + xy

**Simplify each of the following algebraic expressions by removing grouping symbols. 3 x + 2y âˆ’ {x âˆ’ (2y âˆ’ 3)}**

We have

3x + 2y âˆ’ {x âˆ’ (2y âˆ’ 3)}

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

Therefore,

= 3x + 2y âˆ’ {x âˆ’ 2y + 3}

= 3x + 2y âˆ’ x + 2y - 3

= 2x + 4y - 3

**Simplify each of the following algebraic expressions by removing grouping symbols. 5 a âˆ’ {3a âˆ’ (2 âˆ’ a) + 4}**

We have

5a âˆ’ {3a âˆ’ (2 âˆ’ a) + 4}

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

Therefore,

= 5a âˆ’ {3a âˆ’ 2 + a + 4}

= 5a âˆ’ 3a + 2 - a - 4

= 5a - 4a - 2

= a - 2

**Simplify each of the following algebraic expressions by removing grouping symbols. a âˆ’ [b âˆ’ {a âˆ’ (b âˆ’ 1) + 3a}]**

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

**Simplify each of the following algebraic expressions by removing grouping symbols. a âˆ’ [2b âˆ’ {3a âˆ’ (2b âˆ’ 3c)}]**

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