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Page 1
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 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
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 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
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 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
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Page 2
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 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
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 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
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 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
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Given - 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
On rearranging,
= -2x
2
- 3x
2
+ 2y
2
- 3y
2
- 2xy + 3xy
On simplifying, we get
= -5x
2
- y
2
+ xy
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x – (2y – 3)}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
Given 5a – {3a – (2 – a) + 4}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
Given a – [b – {a – (b – 1) + 3a}]
First we have to remove the parentheses, then the curly brackets, and then the square
Page 3
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 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
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 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
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 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
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Given - 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
On rearranging,
= -2x
2
- 3x
2
+ 2y
2
- 3y
2
- 2xy + 3xy
On simplifying, we get
= -5x
2
- y
2
+ xy
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x – (2y – 3)}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
Given 5a – {3a – (2 – a) + 4}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
Given a – [b – {a – (b – 1) + 3a}]
First we have to remove the parentheses, then the curly brackets, and then the square
brackets.
Then we get,
= 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]
On simplifying, we get
= a – 2b + 4a + 1
= 5a – 2b + 1
8. a – [2b – {3a – (2b – 3c)}]
Solution:
Given a – [2b – {3a – (2b – 3c)}]
First we have to remove the parentheses, then the braces, and then the square
brackets.
Then we get,
= a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
= a – [2b – 3a + 2b – 3c]
= a – [4b – 3a – 3c]
On simplifying we get,
= a – 4b + 3a + 3c
= 4a – 4b + 3c
9. -x + [5y – {2x – (3y – 5x)}]
Solution:
Given -x + [5y – {2x – (3y – 5x)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – x + [5y – {2x – (3y – 5x)}]
= – x + [5y – {2x – 3y + 5x)]
= – x + [5y – {7x – 3y}]
= – x + [5y – 7x + 3y]
Page 4
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 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
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 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
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 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
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Given - 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
On rearranging,
= -2x
2
- 3x
2
+ 2y
2
- 3y
2
- 2xy + 3xy
On simplifying, we get
= -5x
2
- y
2
+ xy
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x – (2y – 3)}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
Given 5a – {3a – (2 – a) + 4}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
Given a – [b – {a – (b – 1) + 3a}]
First we have to remove the parentheses, then the curly brackets, and then the square
brackets.
Then we get,
= 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]
On simplifying, we get
= a – 2b + 4a + 1
= 5a – 2b + 1
8. a – [2b – {3a – (2b – 3c)}]
Solution:
Given a – [2b – {3a – (2b – 3c)}]
First we have to remove the parentheses, then the braces, and then the square
brackets.
Then we get,
= a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
= a – [2b – 3a + 2b – 3c]
= a – [4b – 3a – 3c]
On simplifying we get,
= a – 4b + 3a + 3c
= 4a – 4b + 3c
9. -x + [5y – {2x – (3y – 5x)}]
Solution:
Given -x + [5y – {2x – (3y – 5x)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – x + [5y – {2x – (3y – 5x)}]
= – x + [5y – {2x – 3y + 5x)]
= – x + [5y – {7x – 3y}]
= – x + [5y – 7x + 3y]
= – x + [8y – 7x]
On simplifying we get
= – x + 8y – 7x
= – 8x + 8y
10. 2a – [4b – {4a – 3(2a – b)}]
Solution:
Given 2a – [4b – {4a – 3(2a – b)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= 2a – [4b – {4a – 3(2a – b)}]
= 2a – [4b – {4a – 6a + 3b}]
= 2a – [4b – {- 2a + 3b}]
= 2a – [4b + 2a – 3b]
= 2a – [b + 2a]
On simplifying, we get
= 2a – b – 2a
= – b
11. -a – [a + {a + b – 2a – (a – 2b)} - b]
Solution:
Given -a – [a + {a + b – 2a – (a – 2b)} - b]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – 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]
On simplifying, we get
= – a + a – 2b
= – 2b
Page 5
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 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
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 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
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 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
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Given - 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
On rearranging,
= -2x
2
- 3x
2
+ 2y
2
- 3y
2
- 2xy + 3xy
On simplifying, we get
= -5x
2
- y
2
+ xy
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x – (2y – 3)}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
Given 5a – {3a – (2 – a) + 4}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
Given a – [b – {a – (b – 1) + 3a}]
First we have to remove the parentheses, then the curly brackets, and then the square
brackets.
Then we get,
= 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]
On simplifying, we get
= a – 2b + 4a + 1
= 5a – 2b + 1
8. a – [2b – {3a – (2b – 3c)}]
Solution:
Given a – [2b – {3a – (2b – 3c)}]
First we have to remove the parentheses, then the braces, and then the square
brackets.
Then we get,
= a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
= a – [2b – 3a + 2b – 3c]
= a – [4b – 3a – 3c]
On simplifying we get,
= a – 4b + 3a + 3c
= 4a – 4b + 3c
9. -x + [5y – {2x – (3y – 5x)}]
Solution:
Given -x + [5y – {2x – (3y – 5x)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – x + [5y – {2x – (3y – 5x)}]
= – x + [5y – {2x – 3y + 5x)]
= – x + [5y – {7x – 3y}]
= – x + [5y – 7x + 3y]
= – x + [8y – 7x]
On simplifying we get
= – x + 8y – 7x
= – 8x + 8y
10. 2a – [4b – {4a – 3(2a – b)}]
Solution:
Given 2a – [4b – {4a – 3(2a – b)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= 2a – [4b – {4a – 3(2a – b)}]
= 2a – [4b – {4a – 6a + 3b}]
= 2a – [4b – {- 2a + 3b}]
= 2a – [4b + 2a – 3b]
= 2a – [b + 2a]
On simplifying, we get
= 2a – b – 2a
= – b
11. -a – [a + {a + b – 2a – (a – 2b)} - b]
Solution:
Given -a – [a + {a + b – 2a – (a – 2b)} - b]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – 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]
On simplifying, we get
= – a + a – 2b
= – 2b
12. 2x – 3y – [3x – 2y -{x – z – (x – 2y)}]
Solution:
Given 2x – 3y – [3x – 2y -{x – z – (x – 2y)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= 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]
On simplifying, we get
= 2x – 3y – 3x + 4y – z
= - x + y – z
13. 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
Solution:
Given 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= 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
14. x
2
- [3x + [2x - (x
2
- 1)] + 2]
Solution:
Given x
2
- [3x + [2x - (x
2
- 1)] + 2]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
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FAQs on Algebraic Expressions (Exercise 7.4) RD Sharma Solutions - Mathematics (Maths) Class 7
| 1. What are algebraic expressions? |  |
Ans. Algebraic expressions are mathematical expressions that consist of variables, constants, and mathematical operations such as addition, subtraction, multiplication, and division. These expressions are used to represent relationships between quantities and can be simplified or evaluated using specific values for the variables.
| 2. How do you simplify algebraic expressions? | |
Ans. To simplify algebraic expressions, you need to combine like terms by adding or subtracting them. Like terms have the same variables raised to the same powers. You can also use the distributive property to simplify expressions by multiplying and simplifying terms within parentheses. The goal is to simplify the expression as much as possible by removing any unnecessary parentheses or combining like terms.
| 3. What is the difference between an equation and an expression? | |
Ans. An equation is a statement that shows the equality between two expressions. It contains an equal sign (=) and represents a balance or relationship between the two sides of the equation. On the other hand, an expression is a mathematical phrase that may contain variables, constants, and operations but does not have an equal sign. Expressions can be simplified or evaluated, while equations are solved to find the values of the variables that satisfy the equality.
| 4. How do you evaluate algebraic expressions? | |
Ans. To evaluate an algebraic expression, you substitute the given values for the variables and simplify the expression using the order of operations (PEMDAS/BODMAS). Start by replacing each variable with its corresponding value, perform any necessary calculations (such as addition, subtraction, multiplication, and division), and simplify the expression until you obtain a numerical value.
| 5. Can algebraic expressions be solved? | |
Ans. Algebraic expressions cannot be "solved" in the same way as equations because they do not contain an equal sign. However, you can simplify or evaluate algebraic expressions by substituting values for the variables. If you have an equation that contains an algebraic expression, you can solve the equation to find the values of the variables that make the equation true.
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Importance of RD Sharma Solutions: Algebraic Expressions (Exercise 7.4)
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RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Notes
RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Notes offer in-depth insights into the specific topic to help you master it with ease.
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RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Class 7 Questions
The "RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
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The content of RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) is prepared as per the latest Class 7 syllabus.
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