## 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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