Prasanna

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A
• RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B
• RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C

Solve each of the following equations and verify the answer in each case :

Question 1.
Solution:
x + 5 = 12
=> x + 5 – 5 = 12 – 5
(Subtracting 5 from both sides)
=> x = 7
.’. x = 7 is the solution of the given equation.
Check : Substituting x = 7 in the given equation, we get
L.H.S. = 7 + 5 = 12 and R.H.S. = 12
∴When x = 7, we have L.H.S. = R.H.S.

Question 2.
Solution:
x + 3 = – 2
=>x + 3 – 3 = – 2 – 3
(Subtracting 3 from both sides)
=> x = – 5
∴ x = – 5 is the solution of the given equation.
Check : Substituting x = – 5 in the given equation, we get:
L.H.S. = – 5 + 3 = – 2 and R.H.S. = – 2
When x = – 5,
∴we have L.H.S. = R.H.S.

Question 3.
Solution:
x – 7 = 6
=>x – 7 + 7 = 6 + 7
(Adding 7 to both sides)
=> x – 13
So, x = 13 is the solution of the given equation.
Check : Substituting x – 13 in the given equation, we get
L.H.S.= 13 – 7 = 6 and R.H.S. = 6
∴When x = 13, we have L.H.S. = R.H.S.

Question 4.
Solution:
x – 2 = – 5
=> x – 2 + 2 = – 5 + 2
(Adding 2 on both sides)
=> x = – 3
So, x = – 3 is the solution of the given equation.
Check : Substituting x = – 3 in the given equation, we get
L.H.S. = – 3 – 2 = – 5 and R.H.S. = – 5 When x = – 3,
we have
L.H.S. = R.H.S.

Question 5.
Solution:
3x – 5 = 13
=>3x – 5 + 5 = 13 + 5
(Adding 5 on both sides)
=> 3x = 18
=>\(\\ \frac { 3x }{ 3 } \) = \(\\ \frac { 18 }{ 3 } \)
(Dividing both sides by 3)
=> x = 6
x = 6 is the solution of the given equation.
Check : Substituting x = 6 in the given equation, we get
L.H.S. = 3 x 6 – 5 = 18 – 5 = 13 and R.H.S. = 13
∴ When x = 6, we have L.H.S. = R.H.S

Question 6.
Solution:
4x + 7 = 15
=> 4x + 7 – 7 = 15 – 7
(Subtracting 7 from both sides)
=> 4x = 8
=> \(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 8 }{ 4 } \)
(Dividing both sides by 4)
=> x = 2
x = 2 is the solution of the given equation.
Check : Substituting x = 2 in the given equation, we get
L.H.S. = 4 x 2 + 7 = 8 + 7 = 15 and R.H.S. = 15
∴When x = 2, we have L.H.S. = R.H.S.

Question 7.
Solution:
\(\\ \frac { x }{ 5 } \) = 12
=> \(\\ \frac { x }{ 5 } \) x 5 = 12 x 5
(Multiplying both sides by 5)
=> x = 60
x = 60 is the solution of the given equation.
Check : Substituting x = 60 in the given equation, we get
L.H.S. = \(\\ \frac { 60 }{ 5 } \) = 12 and R.H.S. = 12
When x = 60, we have
∴L.H.S. = R.H.S.

Question 8.
Solution:
\(\\ \frac { 3x }{ 5 } \) = 15
=> \(\\ \frac { 3x }{ 5 } \) x \(\\ \frac { 5 }{ 3 } \) = 15 x \(\\ \frac { 5 }{ 3 } \)

Question 9.
Solution:
5x – 3 = x + 17
=> 5x – x = 17 + 3

So. x = 5 is a solution of the given
equation.
Check : Substituting x = 5 in the given
equation, we get
L.H.S. = 5 x 5 – 3 = 25 – 3 = 22
R.H.S. 5 + 17 = 22
∴When x = 5, we have L.H.S. = R.H.S.

Question 10.
Solution:
2x – \(\\ \frac { 1 }{ 2 } \) = 3
=> 2x = 3 + \(\\ \frac { 1 }{ 2 } \)

Question 11.
Solution:
3(x + 6) = 24
=> \(\\ \frac { 3(x+6) }{ 3 } \) = \(\\ \frac { 24 }{ 3 } \)
(Dividing both sides by 3)
x + 6 = 8
=> x = 8 – 6
(Transposing 6 to R.H.S.)
=> x = 2
x = 2 is a solution of the given equation.
Check : Substituting the value of x = 2
in the given equation, we get
L.H.S. = 3(2 + 6 ) = 3 x 8 = 24
and RH.S. = 24
∴When x = 2, we have L.H.S. = R.H.S.

Question 12.
Solution:
6x + 5 = 2x + 17
=> 6x – 2x = 17 – 5
(Transposing 2 x to L.H.S. and 5 to R.H.S.)
=> 4x = 12
=> \(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 12 }{ 4 } \)
(Dividing both sides by 4)
=> x = 3
x = 3 is a solution of the given
equation.
Check : Substituting x = 3 in the given
equation, we get
L.H.S.= 6 x 3 + 5 = 18 + 5 = 23
R.H.S.= 2 x 3 + 17 = 6 + 17 = 23
∴When x = 3, we have L.H.S. = R.H.S.

Question 13.
Solution:
\(\\ \frac { x }{ 4 } \) – 8 = 1
=> \(\\ \frac { x }{ 4 } \) = 1 + 8

R.H.S = 1
∴When x = 36,we have L.H.S. = R.H.S.

Question 14.
Solution:
\(\\ \frac { x }{ 2 } \) = \(\\ \frac { x }{ 2 } \) + 1
=> \(\\ \frac { x }{ 2 } \) – \(\\ \frac { x }{ 3 } \) = 1

Question 15.
Solution:
3(x + 2) – 2(x – 1) = 7
=> 3x + 6 – 2x + 2 = 7
(Removing brackets)
3x – 2x + 6 + 2 = 7
x + 8 = 7
x = 7 – 8
(Transposing 8 to R.H.S.)
x = – 1 is a solution of the given
equation.
Check : Substituting x = – 1 in the given
equation, we get
L.H.S. = 3 ( – 1 + 2) – 2( – 1 – 1)
= 3 x 1 + ( – 2 x – 2)
= 3 + 4 = 7 and
R.H.S. = 7
When x = – 1, we have
L.H.S. = R.H.S.

Question 16.
Solution:
5 (x- 1) + 2 (x + 3) + 6 = 0
= 5 (x – 1) + 2 (x + 3) = – 6
(Transposing 6 to R.H.S.)
= 5x – 5 + 2x + 6 = – 6

Question 17.
Solution:
6(1 – 4 x) + 7 (2 + 5 x) – 53
=> 6 – 24x + 14 + 35 x = 53
(Removing brackets)
=> – 24 x + 35 x + 14 + 6 = 53
=> 11 x + 20 = 53
=> 11 x = 53 – 20
=> 11 x = 33
(Transposing 20 to R.H.S.)

Question 18.
Solution:
16 (3x – 5) – 10 (4x – 8) = 40
=> 48x – 80 – 40x + 80 = 40
(Removing brackets)
=> 48x – 40 x – 80 + 80 = 40
=> 8x = 40

Question 19.
Solution:
3 (x + 6) + 2 (x + 3) = 64
=> 3x + 18 + 2x + 6 = 64
(Removing brackets)
=> 3x + 2x + 18 + 6 = 64
=> 5x + 24 = 64
=> 5x = 64 – 24

Question 20.
Solution:
3(2 – 5x) – 2 (1 – 6x) = 1
=> 6 – 15x – 2 + 12x = 1
(Removing brackets)
=> 6 – 2 – 15x + 12x = 1
=> 4 – 3x = 1
– 3x = 1 – 4

Question 21.
Solution:
\(\\ \frac { n }{ 4 } -5\) = \(\\ \frac { n }{ 6 } \) + \(\\ \frac { 1 }{ 2 } \)
Multiplying each term by 12, the L.C.M. of 4, 6, 2, we get

Question 22.
Solution:
\(\\ \frac { 2m }{ 3 } +8\) = \(\\ \frac { m }{ 2 } -1\)
Multiplying each term by 6, the L.C.M. of 2 and 3, we get
\(\\ \frac { 2m }{ 3 } \) x 6 + 8 x 6 = \(\\ \frac { m }{ 2 } \) x 6 – 1 x 6

Question 23.
Solution:
\(\\ \frac { 2x }{ 5 } \) – \(\\ \frac { 3 }{ 2 } \) = \(\\ \frac { x }{ 2 } +1\)
Multiplying each term by 10, the L.C.M. of 5 and 2, we get

Question 24.
Solution:
\(\\ \frac { x-3 }{ 5 } \) – 2 = \(\\ \frac { 2x }{ 5 } \)
multiplying each term by 5, we get

Question 25.
Solution:
\(\\ \frac { 3x }{ 10 } \) – 4 = 14
=> \(\\ \frac { 3x }{ 10 } \) = 14 + 4

Question 26.
Solution:
\(\\ \frac { 3 }{ 4 } (x-1)\) = (x – 3)

Hope given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A
• RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B
• RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C

Question 1.
Solution:
Let x be the given number, then
(i) 5x = 40
(ii) x + 8 = 15
(iii) 25 – x = 7
(iv) x – 5 = 3
(v) 3x – 5 = 16
(vi) x – 12 = 24
(vii) 19 – 2x = 11
(viii) \(\\ \frac { x }{ 8 } \) = 7
(ix) 4x – 3 = 17
(x) 6x = x + 5

Question 2.
Solution:
(i) 7 less than from the number x is 14.
(ii) Twice the number y is 18.
(iii) 11 increased by thrice the number x is 17.
(iv) 3 less than twice the number x is 13.
(v) 30 less than 12 times the number is 6.
(vi) Quotient of twice the number z and 3 is

Question 3.
Solution:
(i) The given equation is 3x – 5 = 7
Substituting x = 4, we get
L.H.S. = 3 x – 5
= 3 x 4 – 5
= 12 – 5
= 7 = R.H.S.
It is verified that x = 4 is the root of the given equation.
(ii) The given equation is 3 + 2 x = 9
Substituting x = 3, we get L.H.S. = 3 + 2x
= 3 + 2 x 3
= 3 + 6 = 9
= R.H.S.
It is verified that x = 3 is the root of the given equation.
(iii) The given equation is 5x – 8 = 2x – 2
Substituting x = 2, we get
L.H.S. = 5x – 8
=5 x 2 – 8
= 10 – 8
= 2
R.H.S. = 2x – 2
= 2 x 2 – 2
= 4 – 2
= 2
L.H.S. = R.H.S.
Hence, it is verified that x = 2, is the root of the given equation.
(iv) The given equation is 8 – 7y = 1 Substituting y = 1, we get L.H.S. = 8 – 7y
= 8 – 7 x 1
= 8 – 7
= 1
= R.H.S.
Hence, it verified that y = 1 is the root of the given equation.
(v) The given equation is \(\\ \frac { z }{ 7 } \) = 8
Substituting the value of z = 56, we get
L.H.S.= \(\\ \frac { 56 }{ 7 } \)
= 8
= R.H.S.
Hence, it is verified that z = 56 is the root of the given equation.

Question 4.
Solution:
(i) The given equation is y + 9 = 13
We try several values of y and find L.H.S. and the R.H.S. and stop when L.H.S. = R.H.S.

When y = 4, we have L.H.S. = R.H.S.
So y = 4 is the solution of the given equation.
(ii) The given equation is x – 1 = 10
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 17, we hive L.H.S. = R.H.S
So x = 17 is the solution of the given equation.
(iii) The given equation is 4x = 28
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 7, we have L.H.S. = R.H.S.
So x = 7 is the solution of the given equation.
(iv) The given equation is 3y = 36
We guess and try several values of y to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When y = 12, we have L.H.S. = R.H.S.
So y = 12 is the solution of the given equation.
(v) The given equation is 11 + x = 19
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 8, we have L.H.S. = R.H.S.
So, x = 8 is the solution of the given equation.
(vi) The given equation is \(\\ \frac { x }{ 3 } \) = 4
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 12, we have L.H.S. = R.H.S.
So, x = 12 is the solution of the given equation.
(vii) The given equation is 2 x – 3 = 9
We guess and try several values of x to find L.H.S. and R.H.S. and stop when

.’. When x = 6, we have L.H.S. = R.H.S.
So, x = 6 is the solution of the given equation.
L.H.S. = R.H.S.
(viii) The given equation is \(\\ \frac { 1 }{ 2 } \) x + 7 = 11
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 8, we have L.H.S. = R.H.S.
So, x = 8 is the solution of the given equation.
(ix) The given equation is 2y + 4 = 3y (x)
We guess and try several values of z to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When y = 4, we have L.H.S. = R.H.S. So, y = 4 is the solution of the given equation
(x) The given equation is z – 3 = 2z – 5
We guess and try several values of z to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S

When z = 2, we have L.H.S. = R.H.S. So, z = 2 is the solution of the given equation.

Hope given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 8 Algebraic Expressions Ex 8D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8A
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8B
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8C
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D

Simplify :

Question 1.
Solution:
We have : a – (b – 2a)
= a – b + 2a
= a + 2a – b
= (1 + 2) a – b
= 3a – b.

Question 2.
Solution:
We have : 4x – (3y – x + 2z)
= 4x – 3y + x – 2z
= 4x + x – 2y – 2z
= 5x – 3y – 2z

Question 3.
Solution:
We have :
(a² + b² + 2ab) – (a² + b² – 2ab)
= a² + b² + 2ab – a² – b² + 2ab
= a² – a² + b² – b² + 2ab + 2ab
= 0 + 0 + (2 + 2) ab
= 4 ab

Question 4.
Solution:
We have :
– 3 (a + b) + 4 (2a – 3b) – (2a – b)
= – 3a – 3b + 8a – 12b – 2a + b
= – 3a + 8a – 2a – 3b – 12b + b
= ( – 3 + 8 – 2) a + ( – 3 – 12 + 1) b
= 3a – 14 b.

Question 5.
Solution:
We have :
– 4x² + {(2x² – 3) – (4 – 3x²)}
= – 4x² + {2x² – 3 – 4 + 3x²}
[removing grouping symbol]
= – 4x² + {5x² – 7)
= – 4x² + 5x² – 7
(removing grouping symbol {})
= x² – 7

Question 6.
Solution:
We have :
– 2 (x² – y² + xy) – 3 (x² + y² – xy)
= – 2x² + 2y² – 2xy – 3x² – 3y² + 3xy
= – 2x² – 3x² + 2y² – 3y² – 2xy + 3xy
= ( – 2 – 3)x² + (2 – 3) y² + ( – 2 + 3)xy
= – 5x² – y² + xy

Question 7.
Solution:
a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
[removing grouping symbol( )]
= a – [2b – 3a + 2b – 3c]
(removing grouping symbol {})
= a – [4b – 3a – 3c]
= a – 4b + 3a + 3c
(removing grouping symbol [ ])
= 4a – 4b + 3c

Question 8.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
– x + [5y – {x – (5y – 2x)}]
= – x + [5y – {x – 5y + 2x}]
= – x + [5y – {3x – 5y}]
= – x + [5y – 3x + 5y]
= – x + [ 10y – 3x]
= – x + 10y – 3x
= – x – 3x + 10y
= – 4x + 10y

Question 9.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
86 – [15x – 7 (6x – 9) – 2 {10x – 5(2 – 3x)}]
= 86 – [15x – 42x + 63 – 2 {10x – 10 + 15x}
= 86 – [ 15x – 42x + 63 – 2 {25x – 10}]
= 86 – [15x – 42x + 63 – 50x + 20]
= 86 – 15x + 42x – 63 + 50x – 20
= (86 – 63 – 20) – 15x + 42x + 50x
= (86 – 83) + (- 15 + 42 + 50) x
= 3 + 77x

Question 10.
Solution:
Removing the innermost grouping ‘ symbol () first, then { } and then [ ], we have :
12x – [3x³ + 5x² – {7x² – (4 – 3x – x³) + 6x³} – 3x]
= 12x – [3x³ – 5x² – {7x² – 4 + 3x + x³ + 6x³} – 3x]
= 12x – [3x³ + 5x² – {7x² – 4 + 3x + 7x³} – 3x]
= 12x – [3x³ + 5x² – 7x² + 4 – 3x – 7x³ – 3x]
= 12x – [3x³ – 7x³ + 5x² – 7x² + 4 – 3x – 3x]
= 12x – [ – 4x³ + 2x² + 4 – 6x]
= 12x + 4x³ + 2x² – 4 + 6x
= 12x + 6x + 4x³ + 2x² – 4
= 18x + 4x³ + 2x² – 4
= 4x³ + 2x² + 18x – 4

Question 11.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
5a – [a² – {2a (1 – a + 4a²) – 3a (a² – 5a – 3)}] – 8a
= 5a – [a² – {2a – 2a² + 8a³ – 3a³ + 15a² + 9a}] – 8a
= 5a – [a² – {2a + 9a – 2a² + 15a² + 8a³ – 3a³}] – 8a
= 5a – [a² – {11a + 13a² + 5a³}] – 8a
= 5a – [a² – 11a – 13a² – 5a³] – 8a
= 5a – a² + 11a + 13a² + 5a³ – 8a
= 5a + 11a – 8a – a² + 13a² + 5a³
= 8a + 12a² + 5a³
= 5a3 + 12a² + 8a.

Question 12.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
3 – [x – {2y – (5x + y – 3) + 2x²} – (x² – 3y)]
= 3 – [x – {2y – 5x – y + 3 + 2x²} – x² + 3y]
= 3 – [x – {y – 5x + 3 + 2x²} – x² + 3y]
= 3 – [x – y + 5x – 3 – 2x² – x² + 3y]
= 3 – [6x + 2y – 3 – 3x²]
= 3 – 6x – 2y + 3 + 3x²
= 6 – 6x – 2y + 3x²

Question 13.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
xy – [yz – zx – {yx – (3y – xz} – (xy – zy)}]
= xy – [yz – zx – {yx – 3y + xz – xy + zy}]
= xv – [yz – zx – {- 3y + xz + zy}]
= xy – [yz – zx + 3y – xz – zy]
= xy – [ – 2xz + 3y]
= xy + 2xz – 3y

Question 14.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
2a – 3b – [3a – 2b – {a – c – (a – 2b)}]
= 2a – 3b – [3a – 2b – {a – c – a + 2b}]
= 2a – 3b – [3a – 2b – { – c + 2b}]
= 2a – 3b – [3a – 2b + c – 2b]
= 2a – 3b – 3a + 2b – c + 2b
= 2a – 3a – 3b + 2b + 2b – c
= – a + b – c

Question 15.
Solution:
Removing the innermost grouping symbol () first, then { } and ten [ ], 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 + 2a – 3b + b
= – 2a + 2a – 2b
= – 2 b

Question 16.
Solution:
Removing the innermost grouping symbol ‘—’ first, then ( ), then { } and then [ ], we have
2a – [4b – {4a – (3b – \(\overline { 2a+2b } \))}]
= 2a – [4b – {4a – (3b – 2a – 2b)}]
= 2a – [4b – {4a – (b – 2a)}]
= 2a – [4b – {4a – b + 2a}]
= 2a – [4b – {6a – b}]
= 2a – [4b – 6a + b]
= 2a – [5b – 6a]
= 2a – 5b + 6a
= 8a – 5b.

Question 17.
Solution:
Removing the innermost grouping < symbol ( ) first, then { } and then [ ], we have :
5x – [4y – {7x – (3z – 2y) + 4z – 3(x + 3y – 2z)}]
= 5x – [4y – {7x – 3z + 2y + 4z – 3x – 9y + 6z}]
= 5x – [4y – {4x + 7z – 7y}]
= 5x – [4y – 4x – 7z + 7y]
= 5x – [11y – 4x – 7z]
= 5x – 11y + 4x + 7z
= 9x – 11y + 7z

Hope given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.