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.

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Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 7 |

Chapter Name | Algebraic Expressions |

Exercise | Ex 7.4 |

Table of Contents

**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(x^{2} – y^{2} + xy) – 3(x^{2} +y^{2} – xy)

**Solution:**

We have

– 2(x^{2} – y^{2} + xy) – 3(x^{2} +y^{2} – 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^{2 }+ 2y^{2} – 2xy – 3x^{2} – 3y^{2} + 3xy

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

= -5x^{2 }– y^{2} + 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:**

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:**

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^{2 }– [3x + [2x – (x^{2 }– 1)] + 2]

**Solution:**

Therefore, we have

x^{2} – [3x + [2x – (x^{2} – 1)] + 2]

= x^{2} – [3x + [2x – x^{2} + 1] + 2]

= x^{2} – [3x + 2x – x^{2} + 1 + 2]

= x^{2} – [5x – x^{2} + 3]

= x^{2} – 5x + x^{2} – 3

= 2x^{2} – 5x – 3

**Question: 15**

Simplify, the algebraic expressions by removing grouping symbols.

20 – [5xy + 3[x^{2} – (xy – y) – (x – y)]]

**Solution:**

20 – [5xy + 3[x^{2} – (xy – y) – (x – y)]]

= 20 – [5xy + 3[x^{2} – xy + y – x + y]]

= 20 – [5xy + 3[x^{2} – xy + 2y – x]]

= 20 – [5xy + 3x^{2} – 3xy + 6y – 3x]

= 20 – [2xy + 3x^{2} + 6y – 3x]

= 20 – 2xy – 3x^{2} – 6y + 3x

= – 3x^{2 }– 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:**

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:**

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**

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