Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions (Part - 2) - Ex - 7.2, Algebraic Expressions, Class 7, Math

RD Sharma Solutions (Part - 2) - Ex - 7.2, Algebraic Expressions, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics PDF Download

QUESTION 9:

Add a3b3 − 3 to the sum of 2a3 − 3b3 − 3ab + 7 and −a3b3 + 3ab −9.

ANSWER 9:

First, we need to find the sum of 2a3 - 3b3 - 3ab + 7 and - a3 + b3 + 3ab - 9.
= (2a3 - 3b3 - 3ab + 7) + (- a3 + b3 + 3ab - 9)
Collecting positive and negative like terms together, we get
= 2a3 - a3- 3b3 + b3 - 3ab + 3ab + 7 - 9
= a3 - 2b3 - 2

Now, the required expression = (a3 + b3 - 3) + (a3 - 2b3 - 2).
                                  = a3 + a3 + b3- 2b3 - 3 - 2
                                  = 2a3 - b3 - 5

QUESTION 10:

Subtract:
 (i) 7a2b from 3a2b
 (ii) 4xy from −3xy

ANSWER 10:

(i) Required expression  =  3a2b - 7a2b
                                        = (3 - 7)a2b
                                        = - 4a2b

(ii) Required expression  = - 3 xy - 4xy
                                         = - 7xy
                                      

QUESTION 11:

Subtract:
 (i) −4x from 3y
 (ii) −2x from −5y

ANSWER 11:

(i) Required expression = (3y) - (-4x)
                                       = 3y + 4x

(ii) Required expression = (-5y) - (-2x)
                                        = -5y + 2x

QUESTION 12:

Subtract:
 (i) 6x3−7x2+5x−3 from 4−5x+6x2−8x36x3-7x2+5x-3 from 4-5x+6x2-8x3
 (ii) −x2−3z from 5x2−y+z+7-x2-3z from 5x2-y+z+7
 (iii) x3+2x2y+6xy2−y3 from y3−3xy2−4x2yx3+2x2y+6xy2-y3 from y3-3xy2-4x2y

ANSWER 12:

(i) Required expression = (4 - 5x + 6x2 - 8x3) - (6x3 - 7x2 + 5x - 3)
                                      = 4 - 5x + 6x2 - 8x3 - 6x3 + 7x2 - 5x + 3
                                      = - 8x3- 6x3 + 7x2 + 6x2- 5x - 5x + 3 + 4
                                      = - 14x3 + 13x2 - 10x +7

(ii) Required expression  = (5x2 - y + z + 7) - (- x2 - 3z)
                                        = 5x2 - y + z + 7 + x2 + 3z
                                        = 5x2+ x2 - y + z + 3z + 7
                                        = 6x2 - y + 4z + 7

(iii) Required expression = (y3 - 3xy2 - 4x2y) - (x3 + 2x2y + 6xy2 - y3)
                                         = y3 - 3xy2 - 4x2y - x3 - 2x2y - 6xy2 + y3
                                         = y3 + y3 - 3xy2- 6xy2 - 4x2y - - x3
                                         = 2y3- 9xy2 - 6x2y - x3

QUESTION 13:

From
 (i) p3 − 4 + 3p2, take away 5p2 − 3p3 + p − 6
 (ii) 7 + xx2, take away 9 + x + 3x2 + 7x3
 (iii) 1 − 5y2, take away y3 + 7y2 + y + 1
 (iv) x3 − 5x2 + 3x + 1, take away 6x2 − 4x3 + 5 + 3x

ANSWER 13:

 (i) Required expression = (p3 - 4 + 3p2) - (5p2 - 3p3 + p - 6)
                                        = p3 - 4 + 3p2 - 5p2 + 3p3 - p + 6
                                        = p3 + 3p3 + 3p2 - 5p2- p - 4+ 6
                                        = 4p3 - 2p2 - p + 2

(ii) Required expression = (7 + x - x2) - (9 + x + 3x2 + 7x3)
                                        = 7 + x - x2 - 9 - x - 3x2 - 7x3
                                        = - 7x3- x2 - 3x2 + 7 - 9
                                        = - 7x3 - 4x2 - 2

(iii) Required expression = (1 - 5y2) - (y3+ 7y2 + y + 1)
                                         = 1 - 5y2 - y3 - 7y2 - y - 1
                                         = - y3- 5y2 - 7y2 - y
                                         = - y3- 12y2 - y

(iv) Required expression = (x3 - 5x2 + 3x + 1) - (6x2 - 4x3 + 5 +3x)
                                         = x3 - 5x2 + 3x + 1 - 6x2 + 4x3 - 5 - 3x
                                         = x3 + 4x3 - 5x2 - 6x2 + 1 - 5
                                         = 5x3 - 11x2 - 4

The document RD Sharma Solutions (Part - 2) - Ex - 7.2, Algebraic Expressions, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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FAQs on RD Sharma Solutions (Part - 2) - Ex - 7.2, Algebraic Expressions, Class 7, Math - RD Sharma Solutions for Class 7 Mathematics

1. What is the importance of algebraic expressions in mathematics?
Ans. Algebraic expressions are important in mathematics as they help in representing relationships and solving problems using symbols and variables. They allow us to generalize patterns, simplify complex equations, and find unknown quantities. Algebraic expressions also play a crucial role in various fields like physics, engineering, and economics.
2. How can I simplify algebraic expressions?
Ans. To simplify algebraic expressions, you need to combine like terms, remove parentheses, and apply the rules of operations like addition, subtraction, multiplication, and division. Start by combining the terms with the same variables and exponents, then simplify the resulting expression further by performing the necessary operations. Practice and familiarity with algebraic rules will help you become proficient in simplifying expressions.
3. What are the different types of algebraic expressions?
Ans. There are several types of algebraic expressions, including: - Monomials: Expressions with only one term, such as 3x or 5y^2. - Binomials: Expressions with two terms, separated by a plus or minus sign, such as 2x + 3y or 4a - 5b. - Trinomials: Expressions with three terms, such as 2x + 3y - 5z. - Polynomials: Expressions with any number of terms, such as 4x^2 + 2xy - 3y^2 or 3a^3 + 2a^2 - a + 5.
4. How can algebraic expressions be used in real-life situations?
Ans. Algebraic expressions have various applications in real-life situations, such as: - Budgeting: Algebraic expressions can help in managing finances by representing income, expenses, and savings. - Geometry: Expressions can be used to calculate the perimeter, area, and volume of shapes. - Physics: Algebraic expressions are used to represent physical laws and formulas for motion, forces, and energy. - Business and Economics: Expressions can be used for profit and loss calculations, interest rates, and investment analysis.
5. How can I solve algebraic expressions?
Ans. To solve algebraic expressions, you need to isolate the variable and find its value. This can be done by applying inverse operations to both sides of the equation. Start by simplifying the expression if necessary, then use addition, subtraction, multiplication, and division to move terms to one side and the variable to the other side. Finally, calculate the value of the variable by performing the required operations.
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