The DAV Maths Book Class 7 Solutions Pdf and **DAV Class 7 Maths Chapter 7 Worksheet 1** Solutions of Linear Equations in One Variable offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 7 WS 1 Solutions

Question 1.

Solve the following equations and check your answers:

(i) 5x – 2 = 18

Answer:

5x – 2 = 18

⇒ 5x – 2 + 2 = 18 + 2

⇒ 5x = 20

⇒ 5x + 5 =20 + 5

⇒ x = 4

Check: L.H.S. = 5 × 4 – 2 = 20 – 2 = 18

R.H.S. = 18

Hence, x = 4 is the solution of the given equation.

(ii) \(\frac{1}{4}\)y + \(\frac{1}{2}\) = 5

Answer:

(iii) \(\frac{3}{5}\)x – 6 = 3

Answer:

R.H.S = 3

Hence, x = 15 is the solution of the given equation.

(iv) 3x + \(\frac{1}{5}\) = 2 – x

Answer:

Hence, x = \(\frac{9}{20}\) is the solution of the given equation.

(v) 8x + 5 = 6x – 5

Answer:

8x + 5 = 6x – 5

⇒ 8x – 6x + 5 = -5

⇒ 2x + 5 = -5

⇒ 2x + 5 – 5 = -5 – 5

⇒ 2x = -10

∴ x = – 5

Check: L.H.S. = 8 × -5 + 5 = -40 + 5

= -35

R.H.S. = 6 × -5 -5 = -30 – 5 = -35

L.H.S. = R.H.S.

Hence, x = – 5 is the solution of the given equation.

(vi) 9z – 13 = 11z + 27

Answer:

9z – 13 = 11z + 27

⇒ 9z – 11z -13 = 27

⇒ -2z -13 = 27

⇒ – 2z -13 + 13 = 27 + 13

⇒ -2z = 40

⇒ z = – 20

Check: L.H.S. = 9 × – 20 -13 = -180 -13

= -193

R.H.S. = 11 × -20 + 27 – = – 220 + 27

= -193

L.H.S. = R.H.S.

Hence, z = – 20 is the solution of the given equation.

(vii) \(\frac{7}{y}\) + 1 = 29

Answer:

= 7 × 4 + 1

= 28 + 1 = 29

R.H.S. = 29

L.H.S. = R.H.S.

Hence, y = \(\frac{1}{4}\) is the solution of the given equation.

(viii) \(\frac{3}{5}\)x + \(\frac{2}{5}\) = 1

Answer:

R.H.S = 1

L.H.S = R.H.S

Hence, x = 1 is the solution of the given equation.

(ix) 4y – 2 = \(\frac{1}{5}\)

Answer:

Hence, y = \(\frac{11}{5}\)

(x) \(\frac{x}{2}\) + \(\frac{x}{4}\) = 12

Answer:

R.H.S = 12

L.H.S = R.H.S

Hence, x = 16 is the solution of the given equation.

(xi) \(\frac{2}{5}\)z = \(\frac{3}{8}\)z + \(\frac{7}{20}\)

Answer:

L.H.S = R.H.S

Hence, z = 14 is the solution of the given equation.

(xii) \(\frac{2}{5}\)y – \(\frac{5}{8}\)y = \(\frac{5}{12}\)

Answer:

L.H.S = R.H.S

Hence, y = \(\frac{-50}{27}\) is the solution of the given equation.

(xiii) 3x + 2 (x + 2) = 20 – (2x – 5)

Answer:

3x + 2 (x + 2) = 20 – (2x – 5)

⇒ 3x + 2x + 4 = 20 – 2x + 5

⇒ 5x + 4 = – 2x + 25

⇒ 5x + 2x = 25 – 4

⇒ 7x = 21

x = 3

Check L.H.S = 3 × 3 + 2 (3 + 2)

= 9 + 2 × 5

= 9 + 10 = 19

R.H.S = 20 – (2 × 3 – 5)

= 20 – (6 – 5)

= 20 – 1 = 19

L.H.S. = R.H.S.

Hence, x = 3 is the solution of the given

(xiv) 13(y – 4) – 3 (y – 9) = 5(y + 4)

Answer:

13(y – 4) – 3 (y – 9) = 5(y + 4)

⇒ 13y – 52 – 3y + 27 = 5y + 20

⇒ 10y – 25 = 5y + 20

⇒ 10y – 5y = 25 + 20

⇒ 5y = 45

⇒ y = 9

Check: L.H.S = 13(9 – 4) – 3 (9 – 9)

= 13 × 5 – 3 × 0

= 65 – 0 = 65

5(9 + 4) = 5 × 13 = 65

L.H.S. = R.H.S.

Hence, y = 9 is the solution of the given equation.

(xv) (2z – 7) – 3 (3z + 8) = 4z – 9

Answer:

(2z – 7) – 3(3z + 8) = 4z – 9

⇒ 2z – 7 – 9z – 24 = 4z – 9

⇒ – 7z – 31 = 4z – 9

⇒ – 7z – 4z = 31 – 9

⇒ – 11z = 22

∴ z = – 2

Check: L.H.S. = (2x – 2 – 7) – 3(3x – 2 + 8)

= (- 4 – 7) – 3 (- 6 + 8)

= -11 – 3(2)

= -11 – 6 = -17

R.H.S. = 4 × -2 – 9 = -8 – 9 = -17

L.H.S. = R.H.S.

Hence, z = – 2 is the solution of the given equation.

(xvi) 4(2y – 3) + 5(3y – 4) = 14

Answer:

4(2y – 3) + 5 (3y – 4) = 14

⇒ 8y -12 + 15y – 20 = 14

⇒ 23y – 32 = 14

⇒ 23y = 14 + 32

⇒ 23y = 46

∴ y = 46 + 23

y = 2

Check: L.H.S. = 4(2 × 2 – 3) + 5 (3 × 2 – 4)

= 4(4 – 3) + 5 (6 – 4)

=4 × 1 + 5 × 2

= 4 + 10

= 14

R.H.S. = 14

L.H.S. = R.H.S.

Hence, y = 2 is the solution of the given equation.

(xvii) \(\frac{x}{2}-\frac{x}{3}=\frac{x}{4}+\frac{1}{2}\)

Answer:

L.H.S = R.H.S

Hemce, x = -6 is the solution of the given equation

(xviii) z – \(\frac{x}{2}-\frac{x}{3}=\frac{x}{4}+\frac{1}{2}\) = 5

Answer:

⇒ \(\frac{6 z-4 z+3 z}{6}\) = 5

⇒ 9z – 4z = 5 × 6

⇒ 5z = 30

⇒ z = 30 ÷ 5 = 6

z = 6

Checks: L.H.S = 6 – \(\frac{2 \times 6}{3}+\frac{6}{2}\) = 6 – 4 +3

= 5 = R.H.S

Hence, z = 6 is the solution of the given equation.

(xix) \(\frac{6 y+1}{2}\) + 1 = \(\frac{7 y-3}{3}\)

Answer:

⇒ \(\frac{6 y+1+2}{2}=\frac{7 y-3}{3}\)

⇒ \(\frac{6 y+3}{2}=\frac{7 y-3}{3}\)

⇒ 3(6y + 3) = 2(7y – 3)

⇒ 18y + 9 = 14y – 6

⇒ 18y – 14y = -9 – 6

⇒ 4y = – 15

L.H.S = R.H.S

Hence, y = \(\frac{-15}{4}\) is the solution of the given equation.

(xx) \(\frac{6 x-2}{5}=\frac{2 x-1}{3}-\frac{1}{3}\)

Answer:

⇒ 3(8x – 1) = – 1 × 15

⇒ 24x – 3 = -15

⇒ 24x = -15 + 3

⇒ 24x = -12

L.H.S = R.H.S

Hence, x = –\(\frac{1}{2}\) is the solution of the given equation.

(xxi) \(\frac{z-1}{3}=1+\frac{z-2}{4}\)

Answer:

⇒ \(\frac{z-1}{3}=\frac{4+z-2}{4}\)

⇒ \(\frac{z-1}{3}=\frac{z+2}{4}\)

⇒ 4(z – 1) = 3(z + 2)

⇒ 4z – 4 = 3z + 6

⇒ 4z – 3z = 4 + 6

⇒ z = 10

Check: L.H.S = \(\frac{10-1}{3}=\frac{9}{3}\) = 3

R.H.S = 1 + \(\frac{10-2}{4}\)

= 1 + \(\frac{8}{4}\)

= 1 + 2 = 3

L.H.S = R.H.S

Hence, z = 10 is the solution of the given equation.

(xxii) 2x – 3 = \(\frac{3}{10}\)(5x – 12)

Answer:

⇒ 10(2x – 3) = 3(5x – 12)

⇒ 20x – 30 = 15x – 36

⇒ 20x – 15x = -36 + 30

⇒ 5x = -6

L.H.S = R.H.S

Hence, x = \(\frac{-6}{5}\) is the solution of the given equation.

(xxiii) 3(y – 3) = 5(2y + 1)

Answer:

⇒ 3y – 9 = 10y + 5

⇒ 3y – 10y = 9 + 5

⇒ – 7y = 14

⇒ y = -2

Check: L.H.S = 3(-2 – 3) = 3 × -5

= -15

R.H.S = 5(2 × -2 + 1) = 5(-4 + 1)

= 5 × – 3

= -15

L.H.S = R.H.S

Hence, y = -2 is the solution of the given equation.

(xxiv) 0.6x + 0.8 = 0.28x + 1.16

Answer:

⇒ 0.6x – 0.28x = 1.16 – 0.8

⇒ 0.32x = 0.36

L.H.S = R.H.S

Hence, x = \(\frac{9}{8}\) is the solution of the given equation.

(xxv) 2.4(3 – x) – 0.6 (2x – 3) = 0

Answer:

⇒ 2.4 × 3 – 2.4x – 0.6 × 2x + 0.6 × 3 = 0

⇒ 7.2 – 2.4x – 1.2x +1.8 = 0

⇒ – 3.6x + 9.0 =0

⇒ -3.6x = -9.0

⇒ x = \(\frac{9}{3.6}=\frac{5}{2}\)

= 2.5

Check: L.H.S. = 2.4 (3 – 2.5) – 0.6 (2 × 2.5 – 3)

= 2.4 × 0.5 – 0.6 (5 – 3)

= 1.20 – 0.6 × 2

= 1.20 – 1.20

= 0 = R.H.S.

L.H.S. = R.H.S.

Hence, x = 2.5 is the solution of the given equation.

### DAV Class 8 Maths Chapter 7 Worksheet 1 Notes

- Equation is a statement of equality having one or more variables.

e.g. 3x – 5 = 0

x^{2}+ 5x – 3 = 0,

y^{2}– 4 = 0,

2x + 3y = 7 etc. - Linear equation in one variable is the equation in which the highest power of the variable is one.

e.g. 3x + 1 = 2x – 3,

\(\frac{5}{6}\) x – \(\frac{1}{3}\) = 4x + 5,

2 (x – 3) = 3 (2x + 5) - The standard form of linear equation in one variable x is ax + b = 0, where a and b are rationals and a ≠ 0.
- The general solution of ax + b = 0 is given by x = – b/a.

**Rules:**

- Adding the same quantity to both sides of an equation, makes no difference.
- Subtracting same quantity from both sides of an equation, makes no difference.
- Multiplying both sides of an equation by the same non-zero quantity, makes no difference.
- Dividing both sides of an equation by the same non-zero quantity makes no difference.
- When a term is transposed from one side to another, its sign is changed.

For example, + changes to – and vice-versa × changes to ÷ and vice-versa.

Finding a solution to a word problem involves following three steps:

- Forming an equation.
- Solving an equation.
- Interpreting the solution.

Example 1:

Solve \(\frac{2 x}{3}\) – 1 = \(\frac{1}{3}\) and check the answer.

Answer:

\(\frac{2 x}{3}\) – 1 = \(\frac{1}{3}\)

\(\frac{2 x}{3}\) – 1 + 1 = \(\frac{1}{3}\) + 1 (adding 1 to bothsides)

\(\frac{2 x}{3}\) = \(\frac{4}{3}\)

\(\frac{2}{3} x \times \frac{3}{2}=\frac{4}{3} \times \frac{3}{2}\) (Multiplying both sides by \(\frac{3}{2}\))

x = 2

Check:

L.H.S. = \(\frac{2}{3}\) × 2 – 1 = \(\frac{4}{3}\) – 1

= \(\frac{4-3}{3}\)

= \(\frac{1}{3}\)

R.H.S. = \(\frac{1}{3}\)

Hence x = 2 is the solution of the given equation.

Example 2:

Solve: 3\(\frac{3}{4}\) x = 5x – 2\(\frac{1}{2}\)

Answer:

3\(\frac{3}{4}\) x = 5x – 2\(\frac{1}{2}\)

⇒ \(\frac{15}{4}\)x = 5x – \(\frac{5}{2}\)

⇒ \(\frac{15}{4}\)x – 5x = \(\frac{-5}{2}\)

(Collecting the like terms on same side)

⇒ \(\frac{15 x-20 x}{4}\) = \(\frac{-5}{2}\)

⇒ \(\frac{-5 x}{4}\) = \(\frac{-5}{2}\)

⇒ \(\frac{-5 x}{4} \times \frac{-4}{5}=\frac{-5}{2} \times \frac{-4}{5}\)

(Multiplying both sides by \(\frac{-4}{5}\))

⇒ x = 2.

Example 3.

Solve: \(\frac{2}{3}\) (x – 3) = 1 – \(\frac{5}{6}\) (3x – 4)

Answer:

\(\frac{2}{3}\) (x – 3) = 1 – \(\frac{5}{6}\) (3x – 4)

⇒ \(\frac{2}{3}\) x – 3 × \(\frac{2}{3}\) = 1 – \(\frac{5}{6}\) × 3x – \(\frac{5}{6}\) × – 4 (Removing the brackets)

⇒ \(\frac{2}{3}\) x – 2 = 1 – \(\frac{5}{2}\) x + \(\frac{10}{3}\)

⇒ \(\frac{2}{3}\) x + \(\frac{5}{2}\) x = 1 + \(\frac{10}{3}\) + 2

(Collecting the like terms on same sides)

⇒ \(\frac{4 x+15 x}{6}=\frac{3+10+6}{3}\)

⇒ \(\frac{19 x}{6}=\frac{19}{3}\)

⇒ \(\frac{19}{6} x \times \frac{6}{19}=\frac{19}{3} \times \frac{6}{19}\)

(Multiplying both sides by \(\frac{6}{19}\))

⇒ x = 2.

Example 4.

On adding 6 to 4 times a number, we get 30. Find the number.

Answer:

Let the required number be x

∴ 4x + 6 = 30

⇒ 4x + 6 – 6 = 30 – 6

⇒ 4x = 24

⇒ x = 6

Hence the required number is 6.

Example 5.

The sum of three consecutive natural numbers is 120. Find the numbers.

Answer:

Let the three natural numbers be x, x + 1, x + 2

∴ x + x + 1 + x + 2 = 120

⇒ 3x + 3 = 120

⇒ 3x + 3 – 3 = 120 – 3

⇒ 3x = 117

⇒ 3x ÷ 3 = 117 ÷ 3

⇒ x = 39

Hence the required numbers are 39, 40 and 41.

Example 6.

There are 48 students in a class. If number of boys are double to the number of girls, find the number of boys in the class.

Answer:

Let the number of girls be x

∴ Number of boys = 2x

∴ x + 2x = 48

⇒ 3x = 48

⇒ 3x ÷ 3 = 48 ÷ 3

⇒ x = 16

Hence the number of boys = 16 × 2 = 32.