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# RD Sharma Solutions (Part - 3) - Ex - 7.2, Algebraic Expressions, Class 7, Math Class 7 Notes | EduRev

## Class 7: RD Sharma Solutions (Part - 3) - Ex - 7.2, Algebraic Expressions, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions (Part - 3) - Ex - 7.2, Algebraic Expressions, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

#### QUESTION 14:

From the sum of 3x2 − 5x + 2 and −5x2 − 8x + 9 subtract 4x2 − 7x + 9.

Required expression = {(3x2 - 5x + 2) + (- 5x2 - 8x + 9)} - (4x2 - 7x + 9)
= {3x2 - 5x + 2 - 5x2 - 8x + 9} -  (4x2 - 7x + 9)
= {3x2 - 5x2 - 5x - 8x + 2 + 9} -  (4x2 - 7x + 9)
= {- 2x2 - 13x +11} - (4x2 - 7x + 9)
= - 2x2 - 13x + 11 - 4x2 + 7x - 9
= - 2x2 - 4x2 - 13x + 7x + 11 - 9
= - 6x2 - 6x + 2

#### QUESTION 15:

Subtract the sum of 13x − 4y + 7z and −6z + 6x + 3y from the sum of 6x − 4y − 4z and 2x + 4y − 7.

Sum of (13x - 4y + 7z) and ( - 6z + 6x + 3y)
= {(13x - 4y + 7z) + (- 6z + 6x + 3y)
={13x - 4y + 7z - 6z + 6x + 3y}
= {13x + 6x - 4y + 3y + 7z - 6z}
= 19x - y + z

Sum of (6x − 4y − 4z) and (2x + 4y − 7)
= (6x − 4y − 4z) + (2x + 4y − 7)
= 6x − 4y − 4z + 2x + 4y − 7
= 8x - 4z - 7

Now, required expression = {(8x - 4z - 7) - (19x - y + z)}
= 8x - 4z - 7 - 19x + y - z
= 8x - 19x + y - 4z - z - 7
= - 11x + y - 5z - 7

#### QUESTION 16:

From the sum of x2 + 3y2 − 6xy, 2x2 − y2 + 8xyy2 + 8 and x2 − 3xy subtract −3x2 + 4y2 − xyx − y + 3.

Sum of (x2 + 3y2 - 6xy), (2x2 - y2 + 8xy), (y2 + 8) and (x2 - 3xy)
={(x2 + 3y2 - 6xy) + (2x2 - y2 + 8xy) + ( y2 + 8) + (x2 - 3xy)}
={x2 + 3y2 - 6xy + 2x2 - y2 + 8xy + y2 + 8 + x2 - 3xy}
= {x2+ 2x2+ x2 + 3y2- y2 + y2- 6xy + 8xy - 3xy + 8}
= 4x2 + 3y2 - xy + 8

Now, required expression = (4x2 + 3y2 - xy + 8) - (- 3x2 + 4y2 - xy + x - y + 3)
= 4x2 + 3y2 - xy + 8 + 3x2 - 4y2 + xy - x + y - 3
= 4x2 + 3x2+ 3y2- 4y2- x + y - 3 + 8
= 7x2 - y2- x + y + 5

#### QUESTION 17:

What should be added to xy − 3yz + 4zx to get 4xy − 3zx + 4yz + 7?

The required expression can be got by subtracting xy - 3yz + 4zx from 4xy - 3zx + 4yz + 7.
Therefore, required expression = (4xy - 3zx + 4yz + 7) - (xy - 3yz + 4zx)
= 4xy - 3zx + 4yz + 7 - xy + 3yz - 4zx
= 4xy - xy - 3zx - 4zx + 4yz + 3yz + 7
= 3xy - 7zx + 7yz + 7

#### QUESTION 18:

What should be subtracted from x2xy + y2x + y + 3 to obtain −x2 + 3y2 − 4xy + 1?

Let 'M' be the required expression. Then, we have
x2 - xy + y2 - x + y + 3 - M = - x2 + 3y2 - 4xy + 1
Therefore,
M = (x2 - xy + y2 - x + y + 3) - (- x2 + 3y2 - 4xy + 1)
= x2 - xy + y2 - x + y + 3 + x2 - 3y2 + 4xy - 1
Collecting positive and negative like terms together, we get
x2 + x2- xy + 4xy + y2- 3y2 - x + y + 3 - 1
= 2x2 + 3xy- 2y2- x + y + 2

#### QUESTION 19:

How much is x − 2y + 3z greater than 3x + 5y − 7?

Required expression  = (x - 2y + 3z) - (3x + 5y - 7)
=  x - 2y + 3z - 3x - 5y + 7
Collecting positive and negative like terms together, we get
x - 3x - 2y - 5y + 3z + 7
= - 2x - 7y + 3z + 7

#### QUESTION 20:

How much is x2 − 2xy + 3y2 less than 2x2 − 3y2 + xy?

Required expression = (2x2 - 3y2 + xy) - (x2 - 2xy + 3y2)
= 2x2 - 3y2 + xy - x2 + 2xy - 3y2
Collecting positive and negative like terms together, we get
2x2 - x2 - 3y2 - 3y2 + xy + 2xy
= x2 - 6y2 + 3xy

#### QUESTION 21:

How much does a2 − 3ab + 2b2 exceed 2a2 − 7ab + 9b2?

Required expression = (a2 - 3ab + 2b2) - (2a2 - 7ab + 9b2)
= a2 - 3ab + 2b2 - 2a2 + 7ab - 9b2
Collecting positive and negative like terms together, we get
=  a- 2a2  - 3ab + 7ab  + 2b2 -  9b2
= - a2 + 4ab - 7b2

#### QUESTION 22:

What must be added to 12x3 − 4x2 + 3x − 7 to make the sum x3 + 2x2 − 3x + 2?

Let 'M' be the required expression. Thus, we have
12x3 - 4x2 + 3x - 7 + M = x3 + 2x2 - 3x + 2
Therefore,
M = (x3 + 2x2 - 3x + 2) - (12x3 - 4x2 + 3x - 7)
=  x3 + 2x2 - 3x + 2 - 12x3 + 4x2 - 3x + 7
Collecting positive and negative like terms together, we get
x3- 12x3 + 2x2 + 4x2 - 3x - 3x + 2 + 7
= - 11x3 + 6x2 - 6x + 9

#### QUESTION 23:

If P = 7x2 + 5xy − 9y2, Q = 4y2 − 3x2 − 6xy and R = −4x2 + xy + 5y2, show that P + Q + R = 0.

We have
P + Q + R = (7x2 + 5xy - 9y2) + (4y2 - 3x2 - 6xy) + (- 4x2 + xy + 5y2)
= 7x2 + 5xy - 9y2 + 4y2 - 3x2 - 6xy - 4x2 + xy + 5y2
Collecting positive and negative like terms together, we get
7x2- 3x2 - 4x+ 5xy - 6xy + xy - 9y2 + 4y2 + 5y2
= 7x2- 7x+ 6xy - 6xy  - 9y2 + 9y
= 0

#### QUESTION 24:

If P = a2b2 + 2ab, Q = a2 + 4b2 − 6ab, R = b2 + b, S = a2 − 4ab and T = −2a2 + b2ab + a. Find P + Q + R + S − T.

We have
P + Q + R + S - T = {(a2 - b2 + 2ab) + (a2 + 4b2 - 6ab) + (b2 + b) + (a2 - 4ab)} - (-2a2 + b2 - ab + a)
= {a2 - b2 + 2ab + a2 + 4b2 - 6ab + b2 + b + a2 - 4ab}- (- 2a2 + b2 - ab + a)
= {3a2 + 4b2 - 8ab + b } - (-2a2 + b2 - ab + a)
= 3a2+ 4b2 - 8ab + b + 2a2 - b2 + ab - a
Collecting positive and negative like terms together, we get
3a2 + 2a2 + 4b2 - b2 - 8ab + ab - a + b
= 5a2 + 3b2- 7ab - a + b

The document RD Sharma Solutions (Part - 3) - Ex - 7.2, 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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## RD Sharma Solutions for Class 7 Mathematics

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