RS Aggarwal 2021-2022 for Class 6 Maths Solutions Chapter 8- Algebraic Expression
RS Aggarwal Class 6 Math Solution Chapter 8- Algebraic Expression Exercise 8C is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 8- Algebraic Expression
Exercise 8C
1. Add:
(i) 3x, 7x
(ii) 7y, - 9y
(iii) 2xy, 5xy, -xy
(iv) 3x, 2y
(v) 2x2, -3x2, 7x2
(vi) 7xyz, −5xyz, 9xyz, −8xyz
(vii) 6a3, −4a3, 10a3, −8a3
(viii) x2−a2, −5x2+2a2, −4x2+4a2
Solution:
(i) Required sum = 3x+7x=10x
(ii) Required sum = 7y+(−9y)
= 7y−9y
= −2y
(iii) Required sum = 2xy+5xy+(−xy)
= 2xy+5xy−xy
= 7xy−xy
= 6xy
(iv) Required sum = 3x+2y
(v) Required sum = 2x2+(−3x2)+7x2
= 2x2−3x2+7x2
= 9x2−3x2
= 6x2
(vi) Required sum = 7xyz+(−5xyz)+9xyz+(−8xyz)
= 7xyz−5xyz+9xyz−8xyz
= 16xyz−13xyz
= 3xyz
(vii) Required sum = 6a3+(−4a3)+10a3+(−8a3)
= 6a3−4a3+10a3−8a3
= 16a3−12a3
= 4a3
(viii) Required sum = x2−a2+(−5x2+2a2)+(−4x2+4a2)
= x2−a2−5x2+2a2−4x2+4a2
= x2−5x2−4x2−a2+2a2+4a2
= x2−9x2−a2+6a2
= −8x2+5a2
2. Add the following:
(i) x−3y−2z
5x+7y−z
−7x−2y+4z̲
(ii) m2−4m+5
−2m2+6m−6
−m2−2m−7̲
(iii) 2x2−3xy+y2
−7x2−5xy−2y2
4x2+xy−6y2̲
(iv) 4xy−5yz−7zx
−5xy+2yz+zx
−2xy−3yz+3zx̲
Solution:
(i) x−3y−2z
5x+7y−z
−7x−2y+4z̲
−x+2y+z̲
(ii) m2−4m+5
−2m2+6m−6
−m2−2m−7̲
−2m2+0−8̲ = −2m2−8
(iii) 2x2−3xy+y2
−7x2−5xy−2y2
4x2+xy−6y2̲
−x2−7xy−7y2̲
(iv) 4xy−5yz−7zx
−5xy+2yz+zx
−2xy−3yz+3zx̲
−3xy−6yz−3zx̲
3. Add:
(i) 3a−2b+5c, 2a+5b−7c, −a−b+c
(ii) 8a−6ab+5b, −6a−ab−8b, −4a+2ab+3b
(iii) 2x3−3x2+7x−8, −5x3+2x2−4x+1, 3−6x+5x2−x3
(iv) 2x2−8xy+7y2−8xy2, 2xy2+6xy−y2+3x2, 4y2−xy−x2+xy2
(v) x3+y3−z3+3xyz, −x3+y3+z3−6xyz, x3−y3−z3−8xyz
(vi) 2+x−x2+6x3, −6−2x+4x2−3x3, 2+x2, 3−x3+4x−2x2
Solution:
(i) (3a−2b+5c)+(2a+5b−7c)+(−a−b+c)
= 3a+2a−a−2b+5b−b+5c−7c+c
= 5a−a−3b+5b+6c−7c
= 4a+2b−c
(ii) (8a−6ab+5b)+(−6a−ab−8b)+(−4a+2ab+3b)
= 8a−6ab+5b−6a−ab−8b−4a+2ab+3b
= 8a−6a−4a−6ab−ab+2ab+5b+3b−8b
= 8a−10a−7ab+2ab+8b−8b
= −2a−5ab
(iii) (2x3−3x2+7x−8)+(−5x3+2x2−4x+1)+(3−6x+5x2−x3)
= 2x3−3x2+7x−8−5x3+2x2−4x+1+3−6x+5x2−x3
= 2x3−5x3−x3−3x2+2x2+5x2+7x−4x−6x−8+1+3
= 2x3−6x3−3x2+7x2+7x−10x−8+4
= −4x3+4x2−3x−4
(iv) (2x2−8xy+7y2−8xy2)+(2xy2+6xy−y2+3x2)+(4y2−xy−x2+xy2)
= 2x2−8xy+7y2−8xy2+2xy2+6xy−y2+3x2+4y2−xy−x2+xy2
= 2x2+3x2−x2+7y2+4y2−y2−8xy+6xy−xy−8xy2+2xy2+xy2
= 5x2−x2+11y2−y2−9xy+6xy−8xy2+3xy2
= 4x2+10y2−3xy−5xy2
(v) (x3+y3−z3+3xyz)+(−x3+y3+z3−6xyz)+(x3−y3−z3−8xyz)
= x3+y3−z3+3xyz−x3+y3+z3−6xyz+x3−y3−z3−8xyz
= x3−x3+x3+y3+y3−y3−z3+z3−z3+3xyz−6xyz−8xyz
= x3+y3−z3+3xyz−14xyz
= x3+y3−z3−11xyz
(vi) (2+x−x2+6x3)+(−6−2x+4x2−3x3)+(2+x2)+(3−x3+4x−2x2)
= 2+x−x2+6x3−6−2x+4x2−3x3+2+x2+3−x3+4x−2x2
= 6x3−3x3−x3−x2−2x2+4x2+x2+x−2x+4x+2−6+2+3
= 6x3−4x3−3x2+5x2+5x−2x+7−6
= 2x3+2x2+3x+1
4. Subtract:
(i) 5x from 2x
(ii)−xy from 6xy
(iii) 3a from 5b
(iv) −7x from 9y
(v) 10x2 from −7x2
(vi) a2−b2 from b2−a2
Solution:
(i) 2x–5x=–3x
(ii) 6xy
= 6xy + xy
= 7xy
(iii) 5b − 3a
(iv) 9y − (− 7x)
= 9y + 7x
(v) − 7x^2 − 10x^2
= − 17x^2
(vi) b^2 − a^2 − (a^2 − b^2)
= b^2 − a^2 − a^2 + b^2
= 2b^2 − 2a^2
5. Subtract:
(i) 5a + 7b − 2c from 3a − 7b + 4c
(ii) a − 2b − 3c from −2a + 5b − 4c
(iii) 5x^2 − 3xy + y^2 from 7x^2 − 2xy − 4y^2
(iv) 6x^3 − 7x^2 + 5x − 3 from 4 − 5x + 6x^2 − 8x^3
(v) x^3 + 2x^2y + 6xy^2 − y^3 from y^3 − 3xy^2 − 4x^2y
(vi) −11x^2y^2 + 7xy − 6 from 9x^2y^2 − 6xy + 9
(vii) −2a + b + 6d from 5a − 2b − 3c
Solution:
(i) (3a − 7b + 4c) − ( 5a + 7b − 2c )
= 3a − 7b + 4c − 5a − 7b + 2c
= 3a − 5a − 7b − 7b + 4c + 2c
= − 2a − 14b + 6c
(ii) (−2a + 5b − 4c) − (a − 2b − 3c)
= −2a + 5b − 4c − a + 2b + 3c
= −2a − a + 5b + 2b − 4c + 3c
= −3a + 7b − c
(iii) (7x^2 − 2xy − 4y^2) − (5x^2 − 3xy + y^2)
= 7x^2 − 2xy − 4y^2 − 5x^2 + 3xy − y^2
= 7x^2 − 5x^2 − 2xy + 3xy − 4y^2 − y^2
= 2x^2 + xy − 5y^2
(iv) (4 − 5x + 6x^2 − 8x^3) − (6x^3 − 7x^2 + 5x − 3)
= 4 − 5x + 6x^2 − 8x^3 − 6x^3 + 7x^2 − 5x + 3
= − 8x^3 − 6x^3 + 6x^2 + 7x^2 − 5x − 5x + 4 + 3
= − 14x^3 + 13x^2 − 10x + 7
(v) (y^3 − 3xy^2 − 4x^2y) − (x^3 + 2x^2y + 6xy^2 − y^3)
= y^3 − 3xy^2 − 4x^2y − x^3 − 2x^2y − 6xy^2 + y^3
= y^3 + y^3 − 3xy^2 − 6xy^2 − 4x^2y − 2x^2y − x^3
= 2y^3 − 9xy^2 − 6x^2y − x^3
(vi) (9x^2y^2 − 6xy + 9) − (−11x^2y^2 + 7xy − 6)
= 9x^2y^2 − 6xy + 9 + −11x^2y^2 − 7xy + 6
= 9x^2y^2 − 11x^2y^2 − 6xy − 7xy + 9 + 6
= − 2x^2y^2 − 13xy + 15
(vii) (5a − 2b − 3c) − (−2a + b + 6d)
= 5a − 2b − 3c + 2a − b − 6d
= 5a + 2a − 2b − b − 3c − 6d
= 7a − 3b − 3c − 6d
6. Simplify:
(i) 2p^3 − 3p^2 + 4p − 5 − 6p^3 + 2p^2 − 8p − 2 + 6p + 8
(ii) 2x^2 − xy + 6x − 4y + 5xy − 4x + 6x^2 + 3y
(iii) x^4 − 6x^3 + 2x − 7 + 7x^3 − x + 5x^2 + 2 − x^4
Solution:
(i) 2p^3 − 3p^2 + 4p − 5 − 6p^3 + 2p^2 − 8p − 2 + 6p + 8
= 2p^3 − 6p^3 − 3p^2 + 2p^2 + 4p − 8p + 6p − 5 − 2 + 8
= − 4p^3 − p^2 + 10p − 8p − 7 + 8
= − 4p^3 − p^2 + 2p + 1
(ii) 2x^2 − xy + 6x − 4y + 5xy − 4x + 6x^2 + 3y
= 2x^2 + 6x^2 − xy + 5xy + 6x − 4x − 4y + 3y
= 8x^2 + 4xy + 2x − y
(iii) x^4 − 6x^3 + 2x − 7 + 7x^3 − x + 5x^2 + 2 − x^4
= cancel (x^4) − cancel (x^4) − 6x^3 + 7x^3 + 5x^2 + 2x − x − 7 + 2
= x^3 + 5x^2 + x − 5
7. From the sum of 3x^2 − 5x + 2 and −5x^2 − 8x + 6, subtract 4x^2 − 9x + 7
Solution:
Sum of 3x^2 − 5x + 2 and −5x^2 − 8x + 6
= (3x^2 − 5x + 2) + (− 5x^2 − 8x + 6)
= 3x^2 − 5x + 2 − 5x^2 − 8x + 6
= 3x^2 − 5x^2 − 5x − 8x + 2 + 6
= − 2x^2 − 13x + 8
Now, subtract 4x^2 − 9x + 7 from − 2x^2 − 13x + 8
= (− 2x^2 − 13x + 8) − (4x^2 − 9x + 7)
= − 2x^2 − 13x + 8 − 4x^2 + 9x − 7
= − 2x^2 − 4x^2 − 13x + 9x + 8 − 7
= − 6x^2 − 4x + 1
8. If A = 7x^2 + 5xy − 9y^2, B = −4x^2 + xy + 5y^2, C = 4y^2 − 3x^2 − 6xy
then show that A + B + C = 0
Solution:
A = cancel (7x^2) + cancel (5xy) − cancel (9y^2)
B = cancel (− 4x^2) + cancel (xy) + cancel (5y^2)
underline (C = cancel (− 3x^2) − cancel (6xy) +cancel (4y^2))
Now, A + B + C = 0
9. What must be add to 5x^3 − 2x^2 + 6x + 7 to make the sum x^3 + 3x^2 − x + 1?
Solution:
Required Number = (x^3 + 3x^2 − x + 1) − (5x^3 − 2x^2 + 6x + 7)
= x^3 + 3x^2 − x + 1 − 5x^3 + 2x^2 − 6x − 7
= x^3 − 5x^3 + 3x^2 + 2x^2 − x − 6x + 1 − 7
= − 4x^3 + 5x^2 − 7x − 6
10. Let P = a^2 − b^2 + 2ab, Q = a^2 + 4b^2 − 6ab, R = b^2 + 6, S = a^2 − 4ab and T = −2a^2 + b^2 − ab + a. Find P + Q + R + S − T.
Solution:
P = a^2 − b^2 + 2ab
Q = a^2 + 4b^2 − 6ab
R = b^2 + 6
S = a^2 − 4ab
T = −2a^2 + b^2 − ab + a
Now, first we add P, Q, R and S
P + Q + R + S
= (a^2 − b^2 + 2ab) + (a^2 + 4b^2 − 6ab) + (b^2 + 6) + (a^2 − 4ab)
= a^2 − b^2 + 2ab + a^2 + 4b^2 − 6ab + b^2 + 6 + a^2 − 4ab
= a^2 + a^2 + + a^2 − b^2 + 4b^2 + b^2 + 2ab − 6ab − 4ab + 6
= 3a^2 − b^2 + 5b^2 + 2ab − 10ab + 6
= 3a^2 + 4b^2 − 8ab + 6
To find P + Q + R + S − T, subtract T = (−2a^2 + b^2 − ab + a)
from P+Q+R+S = (3a^2 + 4b^2 − 8ab + 6)
So, (P + Q + R + S) − (T)
= (3a^2 + 4b^2 − 8ab + 6) − (−2a^2 + b^2 − ab + a)
= 3a^2 + 4b^2 − 8ab + 6 + 2a^2 − b^2 + ab − a
= 3a^2 + 2a^2 + 4b^2 − b^2 − 8ab + ab − a + 6
= 5a^2 + 3b^2 − 7ab − a + 6
11. What must be subtracted from a^3 − 4a^2 + 5a − 6 to obtain a^2 − 2a + 1.
Solution:
Let the required Subtrahend be X
Now,
a^3 − 4a^2 + 5a − 6 − X = a^2 − 2a + 1
⇒ X = a^3 − 4a^2 + 5a − 6 − (a^2 − 2a + 1)
⇒ X = a^3 − 4a^2 + 5a − 6 − a^2 + 2a − 1
⇒ X = a^3 − 4a^2 − a^2 + 5a + 2a − 6 − 1
∴ X = a^3 − 5a^2 + 7a − 7
12. How much is a + 2b − 3c greater than 2a − 3b + c?
Solution:
To get required expression, we have to subtract 2a − 3b + c from a + 2b − 3c.
(a + 2b − 3c ) − (− 2a + 3b − c )
= a + 2b − 3c − 2a + 3b − c
= a − 2a + 2b + 3b − 3c − c
= − a + 5b − 4c
13. How much less than x − 2y + 3z is 2x − 4y − z?
Solution:
To get required expression, we have to subtract 2x − 4y − z from x − 2y + 3z.
(x − 2y + 3z) − (2x − 4y − z)
= x − 2y + 3z − 2x + 4y + z
= x − 2x − 2y + 4y +3z + z
= − x +2y + 4z
14. By how much does 3x^2 − 5x + 6 exceed x^3 − x^2 + 4x − 1 ?
Solution:
To get required expression, we have to subtract x^3 − x^2 + 4x − 1 from 3x^2 − 5x + 6.
(3x^2 − 5x + 6) − (x^3 − x^2 + 4x − 1)
= 3x^2 − 5x + 6 − x^3 + x^2 − 4x + 1
= − x^3 + 3x^2 + x^2 − 5x − 4x + 6 + 1
= − x^3 + 4x^2 − 9x + 7
15. Subtract the sum of 5x − 4y + 6z and −8x + y − 2z from the sum of 12x − y + 3z and −3x + 5y − 8z.
Solution:
First, we add 5x − 4y + 6z and −8x + y − 2z
(5x − 4y + 6z ) + (− 8x + y − 2z)
= 5x − 4y + 6z − 8x + y − 2z
= 5x − 8x − 4y + y + 6z − 2z
= − 3x − 3y + 4z
Now, we add 12x − y + 3z and −3x + 5y − 8z
(12x − y + 3z ) + (−3x + 5y − 8z)
= 12x − y + 3z − 3x + 5y − 8z
= 12x − 3x − y + 5y + 3z − 8z
= 9x + 4y − 5z
Now, we Subtract − 3x − 3y + 4z from 9x + 4y − 5z
(9x + 4y − 5z) − (− 3x − 3y + 4z)
= 9x + 4y − 5z + 3x + 3y − 4z
= 9x + 3x + 4y + 3y − 5z − 4z
= 12x + 7y − 9z
16. By how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2?
Solution:
To get required expression, we have to subtract 2x + 5y − 6z + 2 from 2x − 3y + 4z.
(2x − 3y + 4z) − (2x + 5y − 6z + 2)
= cancel (2x) − 3y + 4z − cancel (2x) − 5y + 6z − 2
= − 3y − 5y + 4z + 6z − 2
= − 8y + 10z − 2
17. By how much does 1 exceed 2x − 3y − 4?
Solution:
To get required expression, we have to subtract 2x − 3y − 4 from 1.
1 − (2x − 3y − 4)`
= 1 − 2x + 3y + 4
= − 2x + 3y + 5
एक टिप्पणी भेजें
एक टिप्पणी भेजें