RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A

RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

Solve the following equations. Check your result in each case.
Question 1.
Solution:
3x – 5 = 0
Adding 5 to both sides
3x – 5 + 5 = 0 + 5
⇒ 3x = 5
⇒ x = \(\frac { 5 }{ 3 }\)
Check:
L.H.S. = 3x – 5
= 3 x \(\frac { 5 }{ 3 }\) – 5
= 5 – 5
= 0
= R.H.S.
Hence x = \(\frac { 5 }{ 3 }\)

Question 2.
Solution:
8x – 3 = 9 – 2x
⇒ 8x + 2x = 9 + 3 (By transposing)
⇒ 10x = 12

Question 3.
Solution:
7 – 5x = 5 – 7x
⇒ – 5x + 7x = 5 – 7 (By transposing)
⇒ 2x = -2
x = -1
Check:
L.H.S. = 7 – 5x = 7 – 5(-1) = 7 + 5 = 12
R.H.S. = 5 – 7x = 5 – 7(-1) = 5 + 7 = 12
L.H.S. = R.H.S.
Hence x = -1

Question 4.
Solution:
3 + 2x = 1 – x
⇒ 2x + x = 1 – 3 (By transposing)
⇒ 3x = -2

Question 5.
Solution:
2(x – 2) + 3(4x – 1) = 0
⇒ 2x – 4 + 12x – 3 = 0
⇒ 2x + 12x = 4 + 3 (By transposing)
⇒ 14x = 7
⇒ x = \(\frac { 7 }{ 14 }\) = \(\frac { 1 }{ 2 }\)
Check : L.H.S. = 2(x – 2) + 3 (4x -1)

Question 6.
Solution:
5 (2x – 3) – 3(3x – 7) = 5
⇒ 10x – 15 – 9x + 21 = 5
⇒ 10x – 9x – 15 + 21 = 5
⇒ 10x – 9x = 5 + 15 – 21 (By transposing)
⇒ x = 20 – 21 = -1
⇒ x = -1
Check:
L.H.S. = 5 (2x – 3) – 3(3x – 7)
= 5[2 x (-1) -3] -3[3 (-1) -7] = 5[-2 – 3] – 3[-3 – 7]
= 5 x (-5) -3 x (-10)
= -25 + 30
= 5 = R.H.S.
Hence x = -1

Question 7.
Solution:

Question 8.
Solution:

L.H.S. = R.H.S.
Hence x = 48

Question 9.
Solution:

Question 10.
Solution:
3x + 2(x + 2) = 20 – (2x – 5)
⇒ 3x + 2x + 4 = 20 – 2x + 5
⇒ 5x + 4 = 25 – 2x
⇒ 5x + 2x = 25 – 4 (By transposing)
⇒ 7x = 21
⇒ x = 3
Check:
L.H.S.= 3x + [2(x + 2)] = 3 x 3 + 2(3 + 2) = 9 + 2 x 5 = 9 + 10 = 19
R.H.S. = 20 – (2x – 5) = 20 – (2 x 3 – 5) = 20 – (6 – 5) = 20 – 1 = 19
L.H.S. = R.H.S.
Hence x = 3

Question 11.
Solution:
13(y – 4) – 3(y – 9) – 5(y + 4) = 0
⇒ 13y – 52 – 3y + 27 – 5y – 20 = 0
⇒ 13y – 3y – 5y – 52 + 27 – 20 = 0
⇒ 13y – 8y – 72 + 27 = 0
⇒ 5y – 45 = 0
⇒ 5y = 45 (By transposing)
⇒ y = 9
Check:
L.H.S. = 13(y – 4) – 3(y – 9) – 5(y + 4)
= 13(9 – 4) – 3(9 – 9) – 5(9 + 4)
= 13 x 5 – 3 x 0 – 5 x 13
= 65 – 0 – 65 = 0 = R.H.S.
Hence y = 9

Question 12.
Solution:
\(\frac { 2m + 5 }{ 3 }\) = 3m – 10
⇒ 2m + 5 = 3 (3m – 10) (By cross multiplication)
⇒ 2m + 5 = 9m – 30
⇒ 2m – 9m = -30 – 5
⇒ -7m = -35
⇒ m = 5
m = 5
Check:

R.H.S. = 3m – 10 = 3 x 5 – 10 = 15 – 10 = 5
L.H.S. = R.H.S.
Hence m = 5

Question 13.
Solution:
6(3x + 2) – 5(6x – 1) = 3(x – 8) – 5(7x – 6) + 9x
⇒ 18x + 12 – 30x + 5 = 3x – 24 – 35x + 30 + 9x
⇒ 18x – 30x + 12 + 5 = 3x – 35x + 9x – 24 + 30
⇒ -12x + 17 = -23x + 6
⇒ – 12x + 23x = 6 – 17
⇒ 11x = -11
x = – 1
Check:
L.H.S. = 6(3x + 2) – 5(6x – 1)
= 6[3x (-1) + 2] – 5[6 x (-1) x -1]
= 6[-3 + 2] – 5[-6 – 1]
= 6 x (-1) – 5 x (-7)
= -6 + 35 = 29
R.H.S. = 3(x – 8) – 5 (7x – 6) + 9x
= 3[-1 – 8] -5 [7 x (-1) – 6] + 9 (-1)
= 3 x (-9) – 5 [-7 – 6] – 9
= -27 – 5(-13) – 9
= -27 + 65 – 9
= 65 – 36 = 29 .
L.H.S. = R.H.S.
Hence x = -1

Question 14.
Solution:
t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t – 4)
⇒ t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12t
⇒ t – 2t + 10t – 10 = 8t – 3t + 18
⇒ 9t – 10 = 5t + 18
⇒ 9t – 5t = 18 + 10 (By transposing)
⇒ 4t = 28
⇒ t = 7
Check:
L.H.S. = t – [2t + 5] -5[1 – 2t]
= 7 – [2 x 7 + 5] – 5[1 – 2 x 7]
= 7 – [14 + 5] – 5 [1 – 14]
= 7 – 19 – 5(-13)
= 7 – 19 + 65
= 72 – 19 = 53
R.H.S. = 2[3 + 4t) – 3(t – 4)
= 2 (3 + 4 x 7) – 3(7 – 4)
= 2(3 + 28) – 3(3)
= 2(31) – 9 = 62 – 9 = 53
L.H.S. = R.H.S.
Hence t = 7 Ans.

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:
0.18 (5x – 4) = 0.5x + 0.8

Question 29.
Solution:
2.4 (3 – x) – 0.6 (2x – 3) = 0
⇒ 7.2 – 2.4x – 1.2x + 1.8 = 0
⇒ -2.4x – 1.2x = – (7.2 + 1.8).
L.H.S. = 2.4 (3 – x) – 0.6 (2x – 3)
⇒ 2.4 (3 – 2.5) – 0.6 (2 x 2.5 – 3)
⇒ 2.4 (0.5) – 0.6 (5 – 3)
⇒ 1.2 – 0.6 x 2 = 1.2 – 1.2 = 0 = R.H.S.
Hence x = 2.5

Question 30.
Solution:
0.5x – (0.8 – 0.2x) = 0.2 – 0.3x
⇒ 0.5x – 0.8 + 0.2x = 0.2 – 0.3x
⇒ 0.5x + 0.2x + 0.3x = 0.2 + 0.8
⇒ 1.0x = 1.0
⇒ x = 1
Check :
L.H.S. = 0.5x – (0.8 – 0.2x)
= 0.5 x 1 – (0.8 – 0.2 x 1)
= 0.5 – (0.8 – 0.2) = 0.5 – 0.6 = -0.1
R.H.S. = 0.2 – 0.3x = 0.2 – 0.3 x 1 = 0.2 – 0.3 = -0.1
L.H.S. = R.H.S.
Hence x = 1

Question 31.
Solution:

Question 32.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A are helpful to complete your math homework.

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