Linear Equations in One Variable

For $\dfrac{x + 0.5}{6x - 1} = \dfrac{1}{2}$ , the value of x is:

1. $\dfrac{1}{2}$

2. $-\dfrac{1}{2}$

3. 2

4. -2

Answer

$\dfrac{x + 0.5}{6x - 1} = \dfrac{1}{2}$

On cross multiplying, we get

⇒ (x + 0.5) $\times$ 2 = 1 $\times$ (6x - 1)

⇒ 2 $\times$ x + 2 $\times$ 0.5 = 1 $\times$ 6x - 1 $\times$ 1

⇒ 2x + 1 = 6x - 1

⇒ 6x - 2x = 1 + 1

⇒ 4x = 2

⇒ x = $\dfrac{2}{4}$

⇒ x = $\dfrac{1}{2}$

Hence, option 1 is the correct option.

For 3(x + 1) - 4x = 0, the value of x is:

1. $\dfrac{1}{3}$

2. $-\dfrac{1}{3}$

3. 3

4. -3

Answer

3(x + 1) - 4x = 0

⇒ 3x + 3 - 4x = 0

⇒ 3 - x = 0

⇒ x = 3

Hence, option 3 is the correct option.

For 7x - 8 = 5x + 2, the value of x is:

1. $\dfrac{5}{6}$

2. $\dfrac{6}{5}$

3. 5

4. 8

Answer

7x - 8 = 5x + 2

⇒ 7x - 5x = 2 + 8

⇒ 2x = 10

⇒ x = $\dfrac{10}{2}$

⇒ x = 5

Hence, option 3 is the correct option.

For $\dfrac{5}{x} = \dfrac{7}{x - 4}$, the value of x is:

1. -10

2. 10

3. 5

4. -5

Answer

$\dfrac{5}{x} = \dfrac{7}{x - 4}$

On cross multiplying, we get :

⇒ 5 $\times$ (x - 4) = 7 $\times$ x

⇒ 5 $\times$ x - 5 $\times$ 4 = 7x

⇒ 5x - 20 = 7x

⇒ 5x - 7x = 20

⇒ - 2x = 20

⇒ x = - $\dfrac{20}{2}$

⇒ x = - 10

Hence, option 1 is the correct option.

If $\dfrac{x - 2}{4} - \dfrac{x - 3}{5} = 1$, the value of x is:

1. -42

2. 42

3. -18

4. 18

Answer

$\dfrac{x - 2}{4} - \dfrac{x - 3}{5} = 1$

Since L.C.M. of denominators 4 and 5 = 20, multiply each term by 20 to get:

⇒ $\dfrac{20(x - 2)}{4} - \dfrac{20(x - 3)}{5} = 1 \times 20$

⇒ 5(x - 2) - 4(x - 3) = 20

⇒ 5x - 10 - 4x + 12 = 20

⇒ x + 2 = 20

⇒ x = 20 - 2

⇒ x = 18

Hence, option 4 is the correct option.

Solve (question no. 2-22) for x :

$\dfrac{3x + 2}{x - 6} = -7$

Answer

$\dfrac{3x + 2}{x - 6} = -7$

On cross multiplying, we get :

⇒ (3x + 2) = - 7 $\times$ (x - 6)

⇒ 3x + 2 = - 7x + 7 $\times$ 6

⇒ 3x + 2 = - 7x + 42

⇒ 3x + 7x = 42 - 2

⇒ 10x = 40

⇒ x = $\dfrac{40}{10}$

⇒ x = 4

Hence, the value of x is 4.

Solve (question no. 2-22) for x :

3a - 4 = 2(4 - a)

Answer

3a - 4 = 2(4 - a)

⇒ 3a - 4 = 8 - 2a

⇒ 3a + 2a = 8 + 4

⇒ 5a = 12

⇒ a = $\dfrac{12}{5}$

⇒ a = 2.4

Hence, the value of a is 2.4.

Solve (question no. 2-22) for x :

3(b - 4) = 2(4 - b)

Answer

3(b - 4) = 2(4 - b)

⇒ 3b - 12 = 8 - 2b

⇒ 3b + 2b = 8 + 12

⇒ 5b = 20

⇒ b = $\dfrac{20}{5}$

⇒ b = 4

Hence, the value of b is 4.

Solve (question no. 2-22) for x :

$\dfrac{x + 2}{9} = \dfrac{x + 4}{11}$

Answer

$\dfrac{x + 2}{9} = \dfrac{x + 4}{11}$

On cross multiplying, we get :

⇒ (x + 2) $\times$ 11 = (x + 4) $\times$ 9

⇒ 11x + 22 = 9x + 36

⇒ 11x - 9x = 36 - 22

⇒ 2x = 14

⇒ x = $\dfrac{14}{2}$

⇒ x = 7

Hence, the value of x is 7.

Solve (question no. 2-22) for x :

$\dfrac{x - 8}{5} = \dfrac{x - 12}{9}$

Answer

$\dfrac{x - 8}{5} = \dfrac{x - 12}{9}$

On cross multiplying , we get :

⇒ (x - 8) $\times$ 9 = (x - 12) $\times$ 5

⇒ 9x - 72 = 5x - 60

⇒ 9x - 5x = - 60 + 72

⇒ 4x = 12

⇒ x = $\dfrac{12}{4}$

⇒ x = 3

Hence, the value of x is 3.

Solve (question no. 2-22) for x :

5(8x + 3) = 9(4x + 7)

Answer

5(8x + 3) = 9(4x + 7)

⇒ 40x + 15 = 36x + 63

⇒ 40x - 36x = 63 - 15

⇒ 4x = 48

⇒ x = $\dfrac{48}{4}$

⇒ x = 12

Hence, the value of x is 12.

Solve (question no. 2-22) for x :

3(x + 1) = 12 + 4(x - 1)

Answer

3(x + 1) = 12 + 4(x - 1)

⇒ 3x + 3 = 12 + 4x - 4

⇒ 3x + 3 = 8 + 4x

⇒ 3x - 4x = 8 - 3

⇒ - x = 5

⇒ x = -5

Hence, the value of x is - 5.

Solve (question no. 2-22) for x :

$\dfrac{3x}{4} - \dfrac{1}{4}(x - 20) = \dfrac{x}{4} + 32$

Answer

$\dfrac{3x}{4} - \dfrac{1}{4}(x - 20) = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{3x}{4} - \dfrac{x}{4} + \dfrac{20}{4} = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{3x}{4} - \dfrac{x}{4} + 5 = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{3x - x}{4} + 5 = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{2x}{4} + 5 = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{2x}{4} - \dfrac{x}{4} = 32 - 5\\[1em] ⇒ \dfrac{2x - x}{4} = 32 - 5\\[1em] ⇒ \dfrac{x}{4} = 27\\[1em] ⇒ x = 27 \times 4\\[1em] ⇒ x = 108$

Hence, the value of x is 108.

Solve (question no. 2-22) for x :

$3a - \dfrac{1}{5} = \dfrac{a}{5} + 5\dfrac{2}{5}$

Answer

$3a - \dfrac{1}{5} = \dfrac{a}{5} + 5\dfrac{2}{5}\\[1em] ⇒ 3a - \dfrac{1}{5} = \dfrac{a}{5} + \dfrac{27}{5}\\[1em] ⇒ 3a - \dfrac{a}{5} = \dfrac{27}{5} + \dfrac{1}{5}\\[1em] ⇒ \dfrac{15a}{5} - \dfrac{a}{5} = \dfrac{27}{5} + \dfrac{1}{5}\\[1em] ⇒ \dfrac{15a - a}{5} = \dfrac{27 + 1}{5} \\[1em] ⇒ \dfrac{14a}{5} = \dfrac{28}{5} \\[1em] ⇒ a = \dfrac{28 \times 5}{5 \times 14}\\[1em] ⇒ a = \dfrac{140}{70}\\[1em] ⇒ a = 2$

Hence, the value of a is 2.

Solve (question no. 2-22) for x :

$\dfrac{x}{3} - 2\dfrac{1}{2} = \dfrac{4x}{9} - \dfrac{2x}{3}$

Answer

$\dfrac{x}{3} - 2\dfrac{1}{2} = \dfrac{4x}{9} - \dfrac{2x}{3}\\[1em] ⇒ \dfrac{x}{3} - \dfrac{5}{2} = \dfrac{4x}{9} - \dfrac{2x}{3}\\[1em]$

Since L.C.M. of denominators 3, 2 and 9 = 18, multiply each term by 18 to get:

⇒ $\dfrac{18 \times x}{3} - \dfrac{18 \times 5}{2} = \dfrac{18 \times 4x}{9} - \dfrac{18 \times 2x}{3}$

⇒ 6x - 9 $\times$ 5 = 2 $\times$ 4x - 6 $\times$ 2x

⇒ 6x - 45 = 8x - 12x

⇒ 6x + 4x = 45

⇒ 10x = 45

⇒ x = $\dfrac{45}{10}$

⇒ x = 4.5

Hence, the value of x is 4.5.

Solve (question no. 2-22) for x :

$\dfrac{4(y + 2)}{5} = 7 + \dfrac{5y}{13}$

Answer

$\dfrac{4(y + 2)}{5} = 7 + \dfrac{5y}{13}\\[1em] ⇒ \dfrac{4y + 8}{5} = 7 + \dfrac{5y}{13}\\[1em] ⇒ \dfrac{4y}{5} + \dfrac{8}{5} = 7 + \dfrac{5y}{13}\\[1em] ⇒ \dfrac{4y}{5} - \dfrac{5y}{13} = 7 - \dfrac{8}{5}\\[1em]$

Since L.C.M. of denominators 5 and 13 = 65, multiply each term by 65 to get:

$⇒ \dfrac{4y \times 65}{5} - \dfrac{5y \times 65}{13} = 7 \times 65 - \dfrac{8 \times 65}{5}$

⇒ 4y $\times$ 13 - 5y $\times$ 5 = 455 - 8 $\times$ 13

⇒ 52y - 25y = 455 - 104

⇒ 27y = 351

⇒ y = $\dfrac{351}{27}$

⇒ y = 13

Hence, the value of y is 13.

Solve (question no. 2-22) for x :

$\dfrac{a + 5}{6} - \dfrac{a + 1}{9} = \dfrac{a + 3}{4}$

Answer

$\dfrac{a + 5}{6} - \dfrac{a + 1}{9} = \dfrac{a + 3}{4}$

Since L.C.M. of denominators 6, 9 and 4 = 36, multiply each term by 36 to get:

$⇒ \dfrac{36(a + 5)}{6} - \dfrac{36(a + 1)}{9} = \dfrac{36(a + 3)}{4}$

⇒ 6(a + 5) - 4(a + 1) = 9(a +3)

⇒ (6a + 30) - (4a + 4) = (9a + 27)

⇒ 6a + 30 - 4a - 4 = 9a + 27

⇒ 2a + 26 = 9a + 27

⇒ 2a - 9a = 27 - 26

⇒ - 7a = 1

⇒ a = -$\dfrac{1}{7}$

Hence, the value of a is $-\dfrac{1}{7}$.

Solve (question no. 2-22) for x :

$\dfrac{2x - 13}{5} - \dfrac{x - 3}{11} = \dfrac{x - 9}{5} + 1$

Answer

$\dfrac{2x - 13}{5} - \dfrac{x - 3}{11} = \dfrac{x - 9}{5} + 1$

Since L.C.M. of denominators 5 and 11 = 55, multiply each term by 55 to get:

$⇒ \dfrac{(2x - 13) \times 55}{5} - \dfrac{(x - 3) \times 55}{11} = \dfrac{(x - 9) \times 55}{5} + 1 \times 55\\[1em]$

⇒ 11 (2x - 13) - 5 (x - 3) = 11 (x - 9) + 55

⇒ (22x - 143) - (5x - 15) = (11x - 99) + 55

⇒ 22x - 143 - 5x + 15 = 11x - 99 + 55

⇒ 17x - 128 = 11x - 44

⇒ 17x - 11x = 128 - 44

⇒ 6x = 84

⇒ x = $\dfrac{84}{6}$

⇒ x = 14

Hence, the value of x is 14.

Solve (question no. 2-22) for x :

6(6x - 5) - 5(7x - 8) = 12(4 - x) + 1

Answer

6(6x - 5) - 5(7x - 8) = 12(4 - x) + 1

⇒ 36x - 30 - 35x + 40 = 48 - 12x + 1

⇒ 1x + 10 = 49 - 12x

⇒ 1x + 12x = 49 - 10

⇒ 13x = 39

⇒ x = $\dfrac{39}{13}$

⇒ x = 3

Hence, the value of x is 3.

Solve (question no. 2-22) for x :

(x - 5)(x + 3) = (x - 7)(x + 4)

Answer

(x - 5)(x + 3) = (x - 7)(x + 4)

⇒ x(x + 3) - 5(x + 3) = x(x + 4) - 7(x + 4)

⇒ x² + 3x - 5x - 15 = x² + 4x - 7x - 28

⇒ x² - 2x - 15 = x² - 3x - 28

⇒ x² - x² - 2x + 3x = - 28 + 15

⇒ 1x = - 13

Hence, the value of x is - 13.

Solve (question no. 2-22) for x :

(x - 5)² - (x + 2)² = -2

Answer

(x - 5)² - (x + 2)² = -2

Using the formula,

[∵ (x - y)² = x² + y² - 2xy]

And,

[∵ (x + y)² = x² + y² + 2xy]

⇒ x² + 5² - 2 $\times$ x $\times$ 5 - (x² + 2² + 2 $\times$ x $\times$ 2) = -2

⇒ x² + 25 - 10x - (x² + 4 + 4x) = -2

⇒ x² + 25 - 10x - x² - 4 - 4x = -2

⇒ 21 - 14x = -2

⇒ - 14x = -2 - 21

⇒ - 14x = - 23

⇒ x = $\dfrac{23}{14}$

⇒ x = $1\dfrac{9}{14}$

Hence, the value of x is $1\dfrac{9}{14}$.

Solve (question no. 2-22) for x :

(x - 1)(x + 6) - (x - 2)(x - 3) = 3

Answer

(x - 1)(x + 6) - (x - 2)(x - 3) = 3

⇒ x(x + 6) - 1(x + 6) - x(x - 3) + 2(x - 3) = 3

⇒ x² + 6x - 1x - 6 - x² + 3x + 2x - 6 = 3

⇒ x² - x² + 6x - 1x + 3x + 2x - 6 - 6 = 3

⇒ 10x - 12 = 3

⇒ 10x = 3 + 12

⇒ 10x = 15

⇒ x = $\dfrac{15}{10}$

⇒ x = $\dfrac{3}{2}$

⇒ x = $1\dfrac{1}{2}$

Hence, the value of x is $1\dfrac{1}{2}$.

Solve (question no. 2-22) for x :

$\dfrac{3x}{x + 6} - \dfrac{x}{x + 5} = 2$

Answer

$\dfrac{3x}{x + 6} - \dfrac{x}{x + 5} = 2\\[1em]$

Since L.C.M. of denominators (x + 6) and (x + 5) = (x + 6)(x + 5), multiply each term with (x + 6)(x + 5) to get:

$⇒ \dfrac{3x(x + 5)(x + 6)}{(x + 6)} - \dfrac{x(x + 5)(x + 6)}{(x + 5)} = 2(x + 5)(x + 6)$

⇒ 3x(x + 5) - x(x + 6) = 2(x + 6)(x + 5)

⇒ (3x² + 15x) - (x² + 6x) = 2(x² + 6x + 5x + 30)

⇒ 3x² + 15x - x² - 6x = 2(x² + 11x + 30)

⇒ 2x² + 9x = 2x² + 22x + 60

⇒ 2x² - 2x² + 9x - 22x = 60

⇒ -13x = 60

⇒ x = - $\dfrac{60}{13}$

⇒ x = - $4\dfrac{8}{13}$

Hence, the value of x is $- 4\dfrac{8}{13}$.

Solve (question no. 2-22) for x :

$\dfrac{1}{x - 1} + \dfrac{2}{x - 2} = \dfrac{3}{x - 3}$

Answer

$\dfrac{1}{x - 1} + \dfrac{2}{x - 2} = \dfrac{3}{x - 3}$

Since L.C.M. of denominators (x - 1), (x - 2) and (x - 3) = (x - 1)(x - 2)(x - 3), multiply each term with (x - 1)(x - 2)(x - 3) to get :

$⇒ \dfrac{1(x - 1)(x - 2)(x - 3)}{(x - 1)} + \dfrac{2(x - 1)(x - 2)(x - 3)}{(x - 2)} = \dfrac{3(x - 1)(x - 2)(x - 3)}{x - 3}$

⇒ (x - 2)(x - 3) + 2(x - 1)(x - 3) = 3(x - 1)(x - 2)

⇒ (x² - 2x - 3x + 6) + 2(x² - 1x - 3x + 3) = 3(x² - 1x - 2x + 2)

⇒ (x² - 5x + 6) + 2(x² - 4x + 3) = 3(x² - 3x + 2)

⇒ x² - 5x + 6 + 2x² - 8x + 6 = 3x² - 9x + 6

⇒ 3x² - 13x + 12 = 3x² - 9x + 6

⇒ 3x² - 3x² - 13x + 9x = 6 - 12

⇒ - 4x = - 6

⇒ x = $\dfrac{6}{4}$

⇒ x = $\dfrac{3}{2}$

⇒ x = $1\dfrac{1}{2}$

Hence, the value of x is $1\dfrac{1}{2}$.

Solve (question no. 2-22) for x :

$\dfrac{x - 1}{7x - 14} = \dfrac{x - 3}{7x - 26}$

Answer

$\dfrac{x - 1}{7x - 14} = \dfrac{x - 3}{7x - 26}$

By cross multiplying, we get :

⇒ (x - 1)(7x - 26) = (x - 3)( 7x - 14)

⇒ x(7x - 26) - 1(7x - 26) = x( 7x - 14) - 3( 7x - 14)

⇒ 7x² - 26x - 7x + 26 = 7x² - 14x - 21x + 42

⇒ 7x² - 33x + 26 = 7x² - 35x + 42

⇒ 7x² - 7x² - 33x + 35x = 42 - 26

⇒ 2x = 16

⇒ x = $\dfrac{16}{2}$

⇒ x = 8

Hence, the value of x is 8.

Solve (question no. 2-22) for x :

$\dfrac{1}{x - 1} - \dfrac{1}{x} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}$

Answer

$\dfrac{1}{x - 1} - \dfrac{1}{x} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}\\[1em]$

Since L.C.M. of denominators x and (x - 1) = x(x - 1) and L.C.M. of (x + 3) and (x + 4) = (x + 3)(x + 4),

$⇒ \dfrac{1 \times x}{x(x - 1)} - \dfrac{1 \times (x - 1)}{x(x - 1)} = \dfrac{1 \times (x + 4)}{(x + 3) \times (x + 4)} - \dfrac{1 \times (x + 3)}{(x + 4) \times (x + 3)}\\[1em] ⇒ \dfrac{x}{x^2 - x} - \dfrac{x - 1}{x^2 - x} = \dfrac{x + 4}{x^2 + 3x + 4x + 12} - \dfrac{x + 3}{x^2 + 3x + 4x + 12}\\[1em] ⇒ \dfrac{x - (x - 1)}{x^2 - x} = \dfrac{x + 4}{x^2 + 7x + 12} - \dfrac{x + 3}{x^2 + 7x + 12}\\[1em] ⇒ \dfrac{x - x + 1}{x^2 - x} = \dfrac{x + 4}{x^2 + 7x + 12} - \dfrac{x + 3}{x^2 + 7x + 12}\\[1em] ⇒ \dfrac{x - x + 1}{x^2 - x} = \dfrac{(x + 4) - (x + 3)}{x^2 + 7x + 12}\\[1em] ⇒ \dfrac{1}{x^2 - x} = \dfrac{x + 4 - x - 3}{x^2 + 7x + 12}\\[1em] ⇒ \dfrac{1}{x^2 - x} = \dfrac{1}{x^2 + 7x + 12}\\[1em]$

By cross multiplying, we get :

$⇒ x^2 + 7x + 12 = x^2 - x\\[1em] ⇒ x^2 - x^2 + 7x + x = - 12\\[1em] ⇒ 8x = - 12\\[1em] ⇒ x = - \dfrac{12}{8}\\[1em] ⇒ x = - \dfrac{3}{2}\\[1em] ⇒ x = - 1\dfrac{1}{2}$

Hence, the value of x is $- 1\dfrac{1}{2}$.

Solve :

$\dfrac{2x}{3} - \dfrac{x - 1}{6} + \dfrac{7x - 1}{4} = 2\dfrac{1}{6}$. Hence, find the value of 'a', if $\dfrac{1}{a} + 5x = 8.$

Answer

$\dfrac{2x}{3} - \dfrac{x - 1}{6} + \dfrac{7x - 1}{4} = 2\dfrac{1}{6}$

$⇒ \dfrac{2x}{3} - \dfrac{x - 1}{6} + \dfrac{7x - 1}{4} = \dfrac{13}{6}$

Since L.C.M. of denominators 3, 6 and 4 = 12, multiply each term with 12 to get:

$⇒ \dfrac{2x \times 12}{3} - \dfrac{(x - 1) \times 12}{6} + \dfrac{(7x - 1) \times 12}{4} = \dfrac{13 \times 12}{6}$

⇒ 2x $\times$ 4 - (x - 1) $\times$ 2 + (7x - 1) $\times$ 3 = 13 $\times$ 2

⇒ 8x - (2x - 2) + (21x - 3) = 26

⇒ 8x - 2x + 2 + 21x - 3 = 26

⇒ 27x - 1 = 26

⇒ 27x = 26 + 1

⇒ 27x = 27

⇒ x = $\dfrac{27}{27}$

⇒ x = 1

Now, when x = 1

$\dfrac{1}{a} + 5x = 8\\[1em] ⇒ \dfrac{1}{a} + 5 \times 1 = 8\\[1em] ⇒ \dfrac{1}{a} + 5 = 8\\[1em] ⇒ \dfrac{1}{a} = 8 - 5\\[1em] ⇒ \dfrac{1}{a} = 3\\[1em] ⇒ a = \dfrac{1}{3}$

Hence, the value of x is 1 and a is $\dfrac{1}{3}$.

Solve :

$\dfrac{4 - 3x}{5} + \dfrac{7 - x}{3} + 4\dfrac{1}{3} = 0$. Hence, find the value of 'p', if 3p - 2x + 1 = 0.

Answer

$\dfrac{4 - 3x}{5} + \dfrac{7 - x}{3} + 4\dfrac{1}{3} = 0\\[1em] ⇒ \dfrac{4 - 3x}{5} + \dfrac{7 - x}{3} + \dfrac{13}{3} = 0\\[1em]$

Since L.C.M. of denominators 5 and 3 = 15, multiply each term with 15 to get:

$⇒ \dfrac{(4 - 3x) \times 15}{5} + \dfrac{(7 - x) \times 15}{3} + \dfrac{13 \times 15}{3} = 0 \times 15$

⇒ (4 - 3x) $\times$ 3 + (7 - x) $\times$ 5 + 13 $\times$ 5 = 0

⇒ (12 - 9x) + (35 - 5x) + 65 = 0

⇒ 12 - 9x + 35 - 5x + 65 = 0

⇒ 112 - 14x = 0

⇒ 14x = 112

⇒ x = $\dfrac{112}{14}$

⇒ x = 8

Now, when x = 8

3p - 2x + 1 = 0

⇒ 3p - 2 $\times$ 8 + 1 = 0

⇒ 3p - 16 + 1 = 0

⇒ 3p - 15 = 0

⇒ 3p = 15

⇒ p = $\dfrac{15}{3}$

⇒ p = 5

Hence, the value of x is 8 and p is 5.

Solve :

$0.25 + \dfrac{1.95}{x} = 0.9$.

Answer

$0.25 + \dfrac{1.95}{x} = 0.9$

⇒ 0.25 $\times$ x + 1.95 = 0.9 $\times$ x

⇒ 0.25x + 1.95 = 0.9x

⇒ 0.25x - 0.9x = - 1.95

⇒ -0.65x = - 1.95

⇒ x = $\dfrac{1.95}{0.65}$

⇒ x = 3

Hence, the value of x is 3.

Solve :

$5x - (4x + \dfrac{5x - 4}{7}) = \dfrac{4x - 14}{3}$

Answer

$5x - (4x + \dfrac{5x - 4}{7}) = \dfrac{4x - 14}{3}\\[1em] ⇒ 5x - 4x - \dfrac{5x - 4}{7} = \dfrac{4x - 14}{3}\\[1em] ⇒ x - \dfrac{5x - 4}{7} = \dfrac{4x - 14}{3}\\[1em] ⇒ \dfrac{x}{1} - \dfrac{5x - 4}{7} = \dfrac{4x - 14}{3}\\[1em]$

Since L.C.M. of denominators 3 and 7 = 21, multiply each term with 21 to get:

$⇒ x \times 21 - \dfrac{(5x - 4)\times 21}{7} = \dfrac{(4x - 14) \times 21}{3}$

⇒ 21x - (5x - 4) $\times$ 3 = (4x - 14) \times 7

⇒ 21x - (15x - 12) = (28x - 98)

⇒ 21x - 15x + 12 = 28x - 98

⇒ 6x + 12 = 28x - 98

⇒ 6x - 28x = - 98 - 12

⇒ - 22x = - 110

⇒ x = $\dfrac{110}{22}$

⇒ x = 5

Hence, the value of x is 5.

The sum of three consecutive even numbers is 90; the middle number is:

1. 28

2. 30

3. 32

4. none of these

Answer

Let the 3 consecutive number be (x - 2), x and (x + 2).

(x - 2) + x + (x + 2) = 90

⇒ x - 2 + x + x + 2 = 90

⇒ 3x = 90

⇒ x = $\dfrac{90}{3}$

⇒ x = 30

Hence, option 2 is the correct option.

The length of a rectangle is 8 m more than its breadth. If the perimeter of the rectangle is 40 m, the length of the rectangle is:

1. 6 m

2. 12 m

3. 14 m

4. 20 m

Answer

Let the breadth of the rectangle be a.

So, length of the rectangle = 8 + a

Perimeter of the rectangle = 40

As we know, perimeter = 2(l + b)

Putting the value,

⇒ 40 = 2((8 + a) + a)

⇒ 40 = 2(8 + a + a)

⇒ 40 = 2(8 + 2a)

⇒ 40 = 16 + 4a

⇒ 4a = 40 - 16

⇒ 4a = 24

⇒ a = $\dfrac{24}{4}$

⇒ a = 6

Length = a + 6 = 8 + 6 = 14 m

Hence, option 3 is the correct option.

Two numbers are in the ratio 5 : 3. If their difference is 18; the numbers are:

1. 27 and -45

2. 45 and -27

3. 45 and 27

4. -45 and -27

Answer

Two numbers are in the ratio 5 : 3

Let the 2 numbers be 5x and 3x.

Difference of two number be 18.

⇒ 5x - 3x = 18

⇒ 2x = 18

⇒ x = $\dfrac{18}{2}$

⇒ x = 9

So, the 2 numbers are 5 x 9 = 45 and 3 x 9 = 27.

Hence, option 3 is the correct option.

Three more than twice a number is equal to four less than the number. The number is:

1. 7

2. -7

3. $\dfrac{1}{7}$

4. $-\dfrac{1}{7}$

Answer

Let the number be x.

So,

3 + 2x = x - 4

⇒ 2x - x = - 4 - 3

⇒ x = - 7

Hence, option 2 is the correct option.

The sum of two numbers is 70 and their difference is 16, the numbers are:

1. 43 and 27

2. 43 and -27

3. -43 and 27

4. -43 and -27

Answer

Let the two numbers be x and y.

So,

x + y = 70

x - y = 16

Adding two equations, we get

⇒ (x + y) + (x - y) = 70 + 16

⇒ x + y + x - y = 70 + 16

⇒ 2x = 86

⇒ x = $\dfrac{86}{2}$

⇒ x = 43

So the other number be y = 70 - x

⇒ y = 70 - 43

⇒ y = 27

Hence, option 1 is the correct option.

Fifteen less than 4 times a number is 9. Find the number.

Answer

Let the number be x.

According to the question,

⇒ 4x - 15 = 9

⇒ 4x = 9 + 15

⇒ 4x = 24

⇒ x = $\dfrac{24}{4}$

⇒ x = 6

Hence, the number be 6.

If Megha's age is increased by three times her age, the result is 60 years. Find her age.

Answer

Let Megha's age be x years.

So,

x + 3x = 60

⇒ 4x = 60

⇒ x = $\dfrac{60}{4}$

⇒ x = 15

Hence, the age of Megha is 15 years.

28 is 12 less than 4 times a number. Find the number.

Answer

Let the number be x.

So,

⇒ 4x - 12 = 28

⇒ 4x = 28 + 12

⇒ 4x = 40

⇒ x = $\dfrac{40}{4}$

⇒ x = 10

Hence, the number be 10.

Five less than 3 times a number is -20. Find the number.

Answer

Let the number be x.

So,

⇒ 3x - 5 = - 20

⇒ 3x = - 20 + 5

⇒ 3x = - 15

⇒ x = - $\dfrac{15}{3}$

⇒ x = - 5

Hence, the number is - 5.

Fifteen more than 3 times Neetu's age is the same as 4 times her age. How old is she ?

Answer

Let the Neetu's age be x years.

⇒ 3x + 15 = 4x

⇒ 4x - 3x = 15

⇒ 1x = 15

Hence, the age of Neetu is 15 years.

A number decreased by 30 is the same as 14 decreased by 3 times the number. Find the number.

Answer

Let the number be x.

So,

⇒ x - 30 = 14 - 3x

⇒ x + 3x = 14 + 30

⇒ 4x = 44

⇒ x = $\dfrac{44}{4}$

⇒ x = 11

Hence, the number is 11.

A's salary is same as 4 times B's salary. If together they earn ₹ 3,750 a month, find the salary of each.

Answer

Let the B's salary be ₹ x.

So, A's salary = ₹ 4x

Therefore,

A's salary + B's salary = 3,750

⇒ 4x + x = 3,750

⇒ 5x = 3,750

⇒ x = $\dfrac{3,750}{5}$

⇒ x = 750

So, A's salary = 4x

= 4 x 750

= 3,000

Hence, A' salary = ₹ 3,000 and B's salary = ₹ 750.

Separate 178 into two parts so that the first part is 8 less than twice the second part.

Answer

Let the first part be x.

So the other part is 178 - x.

⇒ x = 2(178 - x) - 8

⇒ x = 356 - 2x - 8

⇒ x = 348 - 2x

⇒ x + 2x = 348

⇒ 3x = 348

⇒ x = $\dfrac{348}{3}$

⇒ x = 116

Other number = 178 - x

= 178 - 116

= 62

Hence, the two numbers be 116 and 62.

Six more than one-fourth of a number is two fifth of the number. Find the number.

Answer

Let the number be x.

So,

$⇒ 6 + \dfrac{1}{4}x = \dfrac{2}{5}x\\[1em] ⇒ 6 = \dfrac{2}{5}x - \dfrac{1}{4}x$

Since L.C.M. of denominators 5 and 4 = 20, multiply each term with 20 to get:

$⇒ 6 \times 20 = \dfrac{2 \times 20}{5}x - \dfrac{1 \times 20}{4}x$

⇒ 120 = 2 $\times$ 4x - 1 $\times$ 5x

⇒ 120 = 8x - 5x

⇒ 120 = 3x

⇒ x = $\dfrac{120}{3}$

⇒ x = 40

Hence, the number is 40.

The length of a rectangle is twice its width. If its perimeter is 54 cm, find its length.

Answer

Let the breadth of the rectangle be b cm.

The length of a rectangle is twice its width.

l = 2b

Perimeter of the rectangle = 54cm

Perimeter = 2(l + b)

⇒ 2(2b + b) = 54 cm

⇒ 2(3b) = 54 cm

⇒ 6b = 54 cm

⇒ b = $\dfrac{54}{6}$ cm

⇒ b = 9 cm

l = 2b

= 2 x 9

= 18 cm

Hence, length of the rectangle is 18cm.

A rectangle's length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm, the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the original rectangle.

Answer

Let the width of the rectangle be w cm

Length of the rectangle is 5 cm less than twice its width.

l = 2w - 5 cm

New width = old width + 2

= w + 2 cm

New length = old length - 5

= (2w - 5) - 5 cm

= 2w - 5 - 5 cm

= 2w - 10 cm

New perimeter = 2(new length + new width)

⇒ 2((2w - 10) + (w + 2)) = 74 cm

⇒ 2(2w - 10 + w + 2) = 74 cm

⇒ 2(3w - 8) = 74 cm

⇒ 6w - 16 = 74 cm

⇒ 6w = 74 + 16 cm

⇒ 6w = 90 cm

⇒ w = $\dfrac{90}{6}$ cm

⇒ w = 15 cm

Length = 2w - 5 cm

= 2 x 15 - 5 cm

= 30 - 5 cm

= 25 cm

Hence, the length is 25 cm and breadth is 15 cm.

The sum of three consecutive odd numbers is 57. Find the numbers.

Answer

Let the three consecutive odd numbers be (x - 2), x, (x + 2).

So,

(x - 2) + x + (x + 2) = 57

⇒ x - 2 + x + x + 2 = 57

⇒ 3x = 57

⇒ x = $\dfrac{57}{3}$

⇒ x = 19

Other numbers = (x - 2), (x + 2)

= (19 - 2), (19 + 2)

= 17, 21

Hence, the three numbers are 17, 19 and 21.

A man's age is three times that of his son and in twelve years he will be twice as old as his son would be. What are their present ages?

Answer

Let the age of son be x years.

The man's age is three times that of his son, so it equals 3x.

In twelve years, he will be twice as old as his son will be,

Man's age + 12 = 2(son's age + 12)

⇒ 3x + 12 = 2(x + 12)

⇒ 3x + 12 = 2x + 24

⇒ 3x - 2x = 24 - 12

⇒ x = 12

Man's age = 3x

= 3 $\times$ 12

= 36

Hence, the man's age is 36 years and the son's age is 12 years.

A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half the age of the man at that time?

Answer

The age of man = 42 years

The age of son = 12 years

Let the son's age be half of the man's age after x years.

Son's age + x = $\dfrac{1}{2}$ (Man's age + x)

⇒ 12 + x = $\dfrac{1}{2}$ (42 + x)

⇒ 2 $\times$ (12 + x) = 1 $\times$ (42 + x)

⇒ 24 + 2x = 42 + x

⇒ 2x - x = 42 - 24

⇒ x = 18

Hence, after 18 years, the son's age will be half of the man's age.

A man completed a trip of 136 km in 8 hours. Some parts of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.

Answer

Total distance = 136 km

Total time = 8 hours

Let the part of the trip that was covered at 15 km/h be x km.

So, the remaining part, (136 - x)km, was covered at 18km/hr.

As we know, Time = $\dfrac{Distance}{Speed}$

And

Total time = time taken to cover x km at 15 km/h + time taken to cover (136−x) km at 18 km/h.

$⇒ \dfrac{x}{15} + \dfrac{136 - x}{18} = 8\\[1em]$

Since L.C.M. of 15 and 18 = 90, multiply each term with 90 to get:

$⇒ \dfrac{x \times 90}{15} + \dfrac{(136 - x) \times 90}{18} = 8 \times 90$

⇒ x $\times$ 6 + (136 - x) $\times$ 5 = 720

⇒ 6x + 680 - 5x = 720

⇒ x + 680 = 720

⇒ x = 720 - 680

⇒ x = 40

Other part = 136 - x km

= 136 - 40 km

= 96 km

Hence, the man covered 96 km at the speed of 18 km/hr.

The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.

Answer

Let the two numbers be x and y.

So,

x - y = 3 .............(1)

And,

x² - y² = 69

Using the formula

[∵ (x² - y²) = (x - y)(x + y)]

⇒ (x - y)(x + y) = 69

Using eq (1), we get

⇒ 3(x + y) = 69

⇒ x + y = $\dfrac{69}{3}$

⇒ x + y = 23 .............(2)

Adding (1) and (2)

⇒ (x - y) + (x + y) = 3 + 23

⇒ x - y + x + y = 26

⇒ 2x = 26

⇒ x = $\dfrac{26}{2}$

⇒ x = 13

Using eq (2)

⇒ y = 23 - 13

⇒ y = 10

Hence, the numbers are 13 and 10.

Two consecutive natural numbers are such that one-fourth of the smaller exceeds one-fifth of the greater by 1. Find the numbers.

Answer

Let the consecutive natural numbers be x and (x + 1).

So,

$\dfrac{1}{4} \times x - \dfrac{1}{5} \times (x + 1) = 1 \\[1em] ⇒ \dfrac{x}{4} - \dfrac{x + 1}{5} = 1 \\[1em]$

Since L.C.M. of denominators 5 and 4 = 20, multiply each term with 20 to get:

$⇒ \dfrac{x \times 20}{4} - \dfrac{(x + 1) \times 20}{5} = 1\times 20$

⇒ x $\times$ 5 - (x + 1) $\times$ 4 = 20

⇒ 5x - (4x + 4) = 20

⇒ 5x - 4x - 4 = 20

⇒ x - 4 = 20

⇒ x = 20 + 4

⇒ x = 24

Other number = (x + 1)

= 24 + 1

= 25

Hence, the numbers are 24 and 25.

Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively, the sum of the quotients is 40. Find the numbers.

Answer

Let the three consecutive whole numbers be (x - 1), x, (x + 1).

According to the question, the sum of the quotients is 40.

$⇒ \dfrac{x - 1}{5} + \dfrac{x}{3} + \dfrac{x + 1}{4} = 40$

Since L.C.M. of denominators 5, 3 and 4 = 60, multiply each term with 60 to get:

$⇒ \dfrac{(x - 1) \times 60}{5} + \dfrac{x \times 60}{3} + \dfrac{(x + 1) \times 60}{4} = 40 \times 60$

⇒ (x - 1) $\times$ 12 + x $\times$ 20 + (x + 1) $\times$ 15 = 2400\\[1em]

⇒ (12x - 12) + 20x + (15x + 15) = 2400

⇒ 12x - 12 + 20x + 15x + 15 = 40

⇒ 47x + 3 = 2400

⇒ 47x = 2400 - 3

⇒ 47x = 2397

⇒ x = $\dfrac{2397}{47}$

⇒ x = 51

Other number = (x - 1), (x + 1)

= (51 - 1), (51 + 1)

= 50, 52

Hence, the numbers are 50, 51 and 52.

If the same number be added to the numbers 5, 11, 15 and 31, the resulting numbers are in proportion. Find the number.

Answer

Let x be added to each number. The numbers will then be 5 + x, 11 + x, 15 + x and 31 + x.

According to question,

$\dfrac{5 + x}{11 + x} = \dfrac{15 + x}{31 + x}$

By cross multiplying, we get :

⇒ (5 + x)(31 + x) = (15 + x)(11 + x)

⇒ 155 + 5x + 31x + x² = 165 + 11x + 15x + x²

⇒ 155 + 36x + x² = 165 + 26x + x²

⇒ 36x - 26x + x² - x² = 165 - 155

⇒ 10x = 10

⇒ x = $\dfrac{10}{10}$

⇒ x = 1

Hence, 1 is the number that should be added to the given numbers.

The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.

Answer

Let the age of son be x years.

The present age of a man is twice that of his son.

Man's age = 2x

After eight years, their ages will be in the ratio 7 : 4.

$⇒ \dfrac{\text{Man's age after 8 years}}{\text{Son's age after 8 years}} = \dfrac{7}{4}\\[1em] ⇒ \dfrac{\text{Man's present age + 8 years}}{\text{Son's present age + 8 years}} = \dfrac{7}{4}\\[1em] ⇒ \dfrac{2x + 8}{x + 8} = \dfrac{7}{4}\\[1em]$

By cross multiplying, we get

$⇒ 4(2x + 8) = 7(x + 8)\\[1em] ⇒ 8x + 32 = 7x + 56\\[1em] ⇒ 8x - 7x = 56 - 32\\[1em] ⇒ x = 24\\[1em]$

Present son's age = 24 years

Present man's age = 2x years

= 2 $\times$ 24 years

= 48 years

Hence, son's age is 24 years and man's age is 48 years.

For $\dfrac{x}{2} + \dfrac{x}{3} = 15$; the value of x is:

1. 18

2. 90

3. 15

4. 45

Answer

$\dfrac{x}{2} + \dfrac{x}{3} = 15$

Since L.C.M. of denominators 2 and 3 = 6, multiply each term with 6 to get:

$⇒ \dfrac{x \times 6}{2} + \dfrac{x \times 6}{3} = 15 \times 6$

⇒ x $\times$ 3 + x $\times$ 2 = 90

⇒ 3x + 2x = 90

⇒ 5x = 90

⇒ x = $\dfrac{90}{5}$

⇒ x = 18

Hence, option 1 is the correct option.

The sum of three odd numbers is 87. The largest of these is:

1. 27

2. 31

3. 29

4. 25

Answer

Let the three odd numbers be (x - 2), x and (x + 2).

According to question,

⇒ (x - 2) + x + (x + 2) = 87

⇒ x - 2 + x + x + 2 = 87

⇒ 3x = 87

⇒ x = $\dfrac{87}{3}$

⇒ x = 29

Other numbers are (x - 2), (x + 2)

= (29 - 2) , (29 + 2)

= 27 , 31

Hence, option 2 is the correct option.

If $\dfrac{x + 3}{(x + 4) + 3} = \dfrac{4}{5}$, the value of x is:

1. 12

2. -13

3. -12

4. 13

Answer

$\dfrac{x + 3}{(x + 4) + 3} = \dfrac{4}{5}$

⇒ $\dfrac{x + 3}{x + 4 + 3} = \dfrac{4}{5}$

⇒ $\dfrac{x + 3}{x + 7} = \dfrac{4}{5}$

By cross multiplying, we get :

⇒ 5(x + 3) = 4(x + 7)

⇒ 5x + 15 = 4x + 28

⇒ 5x - 4x = 28 - 15

⇒ x = 13

Hence, option 4 is the correct option.

When four consecutive integers are added, their sum is -46. The smallest integer is:

1. 13

2. -13

3. 10

4. -10

Answer

Let the four consecutive integers be (x - 1), x, (x + 1) and (x + 2).

⇒ (x - 1) + x + (x + 1) + (x + 2) = -46

⇒ x - 1 + x + x + 1 + x + 2 = -46

⇒ 4x + 2 = -46

⇒ 4x = -46 - 2

⇒ 4x = -48

⇒ x = - $\dfrac{48}{4}$

⇒ x = - 12

Other numbers are (x - 1), (x + 1), (x + 2)

= (-12 - 1), (-12 + 1), (-12 + 2)

= - 13, - 11, - 10

Hence, option 2 is the correct option.

The perimeter of a rectangular plot is 560 m and the length of the plot is three times its width; the length of the plot is:

1. 140 m

2. 210 m

3. 280 m

4. $186\dfrac{2}{3}m$

Answer

Let the width of the rectangular plot be w.

Length of the rectangular plot = 3w

Perimeter = 560 m

As we know, perimeter of a rectangle = 2(l + w)

⇒ 2 (3w + w) = 560 m

⇒ 2 x 4w = 560 m

⇒ 8w = 560 m

⇒ w = $\dfrac{560}{8}$ m

⇒ w = 70 m

Length of rectangle = 3w

= 3 x 70 m

= 210 m

Hence, option 2 is the correct option.

Statement 1: $\dfrac{ax + b}{cx + d} = \dfrac{p}{q}$.

⇒ q(cx + d) = p(ax + b)

This process is cross-multiplication.

Statement 2: ax + b = c becomes ax = c - b after transposition.

Which of the following options is correct?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, $\dfrac{ax + b}{cx + d} = \dfrac{p}{q}$

On cross-multiplication,

⇒ q(ax + b) = p(cx + d)

So, statement 1 is false.

Transposition is a fundamental algebraic technique used to isolate a variable by moving terms from one side of an equation to the other. When a term is transposed, its sign changes;

• A term added to one side gets subtracted on the other.

• A term subtracted from one side gets added on the other.

∴ ax + b = c becomes ax = c - b after transposition.

So, statement 2 is true.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Assertion (A) : (2x + 5)² - (2x - 5)² = 40 is a linear equation in terms of one variable.

Reason (R) : An equation in which the greatest exponent of the variable after simplification is one is called a linear equation.

1. Both A and R are correct, and R is the correct explanation for A.

2. Both A and R are correct, and R is not the correct explanation for A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

An equation in which the greatest exponent of the variable after simplification is one is called a linear equation.

This definition aligns with the standard definition of a linear equation. A linear equation in one variable is of the form ax + b = 0, where a ≠ 0 and the highest power of x is 1.

So, reason (R) is true.

Given,

⇒ (2x + 5)² - (2x - 5)² = 40

⇒ 4x² + 25 + 20x - (4x² + 25 - 20x) = 40

⇒ 4x² + 25 + 20x - 4x² - 25 + 20x = 40

⇒ 40x = 40

⇒ 40x - 40 = 0

The above equation is in the form ax + b = 0, where a = 40, b = -40 and the highest power of x is 1.

So, assertion (A) is true.

∴ Both A and R are correct, and R is the correct explanation for A.

Hence, option 1 is the correct option.

Assertion (A) : The solution of : $\dfrac{x}{2} + \dfrac{x}{3}$ = 15 is 18.

Reason (R) : In the process of solving a linear equation, only the method of transposition is applied.

1. Both A and R are correct, and R is the correct explanation for A.

2. Both A and R are correct, and R is not the correct explanation for A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Given,

$\Rightarrow \dfrac{x}{2} + \dfrac{x}{3} = 15 \\[1em] \Rightarrow \dfrac{3x + 2x}{6} = 15 \\[1em] \Rightarrow \dfrac{5x}{6} = 15 \\[1em] \Rightarrow x = \dfrac{6 \times 15}{5} \\[1em] \Rightarrow x = \dfrac{90}{5} \\[1em] \Rightarrow x = 18$

So, assertion (A) is true.

In the process of solving a linear equation, multiple methods can be applied, not just transposition.

So, reason (R) is false.

∴ A is true, but R is false.

Hence, option 3 is the correct option.

Assertion (A) : The solution of : $\dfrac{x - 3}{x + 4} = \dfrac{x + 1}{x - 2}$ is 5.

Reason (R) : $\dfrac{ax + b}{cx + d} = \dfrac{p}{q}$.

⇒ q(ax + b) = p(cx + d)

This process is cross-multiplication.

1. Both A and R are correct, and R is the correct explanation for A.

2. Both A and R are correct, and R is not the correct explanation for A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Given, $\dfrac{ax + b}{cx + d} = \dfrac{p}{q}$

On cross-multiplication,

⇒ q(ax + b) = p(cx + d)

So, reason (R) is true.

Given,

$\Rightarrow \dfrac{x - 3}{x + 4} = \dfrac{x + 1}{x - 2} \\[1em] \Rightarrow (x - 3) \times (x - 2) = (x + 4) \times (x + 1) \\[1em] \Rightarrow x(x - 2) - 3(x - 2) = x(x + 1) + 4(x + 1) \\[1em] \Rightarrow x^2 - 2x - 3x + 6 = x^2 + x + 4x + 4 \\[1em] \Rightarrow x^2 - 5x + 6 = x^2 + 5x + 4 \\[1em] \Rightarrow x^2 - 5x + 6 - x^2 - 5x - 4 = 0 \\[1em] \Rightarrow -10x + 2 = 0 \\[1em] \Rightarrow 10x = 2 \\[1em] \Rightarrow x = \dfrac{2}{10} = \dfrac{1}{5}.$

So, assertion (A) is false.

∴ A is false, but R is true.

Hence, option 4 is the correct option.

Assertion (A) : 7x = -21

⇒ $\dfrac{7x}{7} = -\dfrac{21}{7}$ ⇒ x = -3

Reason (R) : An equation remains unchanged on dividing both sides by the same (non-zero) number.

1. Both A and R are correct, and R is the correct explanation for A.

2. Both A and R are correct, and R is not the correct explanation for A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

In algebra, on dividing both sides of an equation by the same non-zero number, the equation remains the same.

So, reason (R) is true.

Given, 7x = -21

Dividing both sides by 7, we get

$\Rightarrow \dfrac{7x}{7} = -\dfrac{21}{7}$

⇒ x = -3

So, assertion (A) is true.

∴ Both A and R are correct, and R is the correct explanation for A.

Hence, option 1 is the correct option.

Solve:

$\dfrac{1}{3}x - 6 = \dfrac{5}{2}$

Answer

$\dfrac{1}{3}x - 6 = \dfrac{5}{2}\\[1em] ⇒ \dfrac{1}{3}x = \dfrac{5}{2} + 6\\[1em] ⇒ \dfrac{1}{3}x = \dfrac{5}{2} + \dfrac{6}{1}$

Since L.C.M. of denominators 3, 2 and 1 = 6, multiply each term with 6 to get:

$⇒ \dfrac{1 \times 6}{3}x = \dfrac{5 \times 6}{2} + \dfrac{6 \times 6}{1}$

⇒ 1 $\times$ 2x = 5 $\times$ 3 + 6 $\times$ 6

⇒ 2x = 15 + 36

⇒ 2x = 51

⇒ x = $\dfrac{51}{2}$

⇒ x = $25\dfrac{1}{2}$

Hence, the value of x is $25\dfrac{1}{2}$.

Solve:

$\dfrac{2x}{3} - \dfrac{3x}{8} = \dfrac{7}{12}$

Answer

$\dfrac{2x}{3} - \dfrac{3x}{8} = \dfrac{7}{12}$

Since L.C.M. of denominators 3 and 8 = 24, multiply each term with 24 to get:

$⇒ \dfrac{2x \times 24}{3} - \dfrac{3x \times 24}{8} = \dfrac{7 \times 24}{12}$

⇒ 2x $\times$ 8 - 3x $\times$ 3 = 7 $\times$ 2

⇒ 16x - 9x = 14

⇒ 7x = 14

⇒ x = $\dfrac{14}{7}$

⇒ x = 2

Hence, the value of x is 2.

Solve:

(x + 2) (x + 3) + (x - 3) (x - 2) - 2x(x + 1) = 0

Answer

(x + 2) (x + 3) + (x - 3) (x - 2) - 2x(x + 1) = 0

⇒ x (x + 3) + 2 (x + 3) + x (x - 2) - 3 (x - 2) - 2x(x + 1) = 0

⇒ x² + 3x + 2x + 6 + x² - 2x - 3x + 6 - 2x² - 2x = 0

⇒ (x² + x² - 2x²) + (3x + 2x - 2x - 3x - 2x) + (6 + 6) = 0

⇒ - 2x + 12 = 0

⇒ 2x = 12

⇒ x = $\dfrac{12}{2}$

⇒ x = 6

Hence, the value of x is 6.

Solve:

$\dfrac{1}{10} - \dfrac{7}{x} = 35$

Answer

$\dfrac{1}{10} - \dfrac{7}{x} = 35\\[1em] ⇒ \dfrac{7}{x} = \dfrac{1}{10} - 35\\[1em] ⇒ \dfrac{7}{x} = \dfrac{1}{10} - \dfrac{35}{1}$

Since L.C.M. of denominators 10 and 1 = 10, multiply each term with 10 to get:

$⇒ \dfrac{7 \times 10}{x} = \dfrac{1 \times 10}{10} - \dfrac{35 \times 10}{1}\\[1em] ⇒ \dfrac{70}{x} = 1 - 350\\[1em] ⇒ \dfrac{70}{x} = - 349\\[1em] ⇒ x = - \dfrac{70}{349}$

Hence, the value of x is $-\dfrac{70}{349}$.

Solve:

13(x - 4) - 3(x - 9) - 5(x + 4) = 0

Answer

13(x - 4) - 3(x - 9) - 5(x + 4) = 0

⇒ 13x - 52 - 3x + 27 - 5x - 20 = 0

⇒ (13x - 3x - 5x) + (- 52 + 27 - 20) = 0

⇒ 5x - 45 = 0

⇒ 5x = 45

⇒ x = $\dfrac{45}{5}$

⇒ x = 9

Hence, the value of x is 9.

Solve:

$x + 7 -\dfrac{8x}{3} = \dfrac{17x}{6} - \dfrac{5x}{8}$

Answer

$x + 7 -\dfrac{8x}{3} = \dfrac{17x}{6} - \dfrac{5x}{8}$

Since L.C.M. of denominators 6 and 8 = 24, multiply each term with 24 to get:

$⇒ x\times 24 + 7\times 24 -\dfrac{8x\times 24}{3} = \dfrac{17x \times 24}{6} - \dfrac{5x \times 24}{8}$

⇒ 24x + 168 - 8x $\times$ 8 = 17x $\times$ 4 - 5x $\times$ 3

⇒ 24x + 168 - 64x = 68x - 15x

⇒ 168 - 40x = 53x

⇒ - 53x - 40x = - 168

⇒ - 93x = - 168

⇒ x = $\dfrac{168}{93}$

⇒ x = $\dfrac{56}{31}$

⇒ x = $1\dfrac{25}{31}$

Hence, the value of x is $1\dfrac{25}{31}$

Solve:

$\dfrac{3x - 2}{4} - \dfrac{2x + 3}{3} = \dfrac{2}{3} - x$

Answer

$\dfrac{3x - 2}{4} - \dfrac{2x + 3}{3} = \dfrac{2}{3} - x\\[1em] ⇒ \dfrac{3x - 2}{4} - \dfrac{2x + 3}{3} + x = \dfrac{2}{3}\\[1em] ⇒ \dfrac{3x - 2}{4} - \dfrac{2x + 3}{3} + \dfrac{x}{1} = \dfrac{2}{3}\\[1em]$

Since L.C.M. of denominators 3 and 4 = 12, multiply each term with 12 to get:

$⇒ \dfrac{12(3x - 2)}{4} - \dfrac{12(2x + 3)}{3} + \dfrac{x \times 12}{1} = \dfrac{2 \times 12}{3}$

⇒ 3(3x - 2) - 4(2x + 3) + x $\times$ 12 = 2 $\times$ 4

⇒ (9x - 6) - (8x + 12) + 12x = 8

⇒ 9x - 6 - 8x - 12 + 12x = 8

⇒ 13x - 18 = 8

⇒ 13x = 8 + 18

⇒ 13x = 26

⇒ x = $\dfrac{26}{13}$

⇒ x = 2

Hence, the value of x is 2.

Solve:

$\dfrac{x + 2}{6} - \Big(\dfrac{11 - x}{3} - \dfrac{1}{4}\Big) = \dfrac{3x - 4}{12}$

Answer

$\dfrac{x + 2}{6} - \Big(\dfrac{11 - x}{3} - \dfrac{1}{4}\Big) = \dfrac{3x - 4}{12}\\[1em] ⇒ \dfrac{x + 2}{6} - \dfrac{11 - x}{3} + \dfrac{1}{4} = \dfrac{3x - 4}{12}$

Since L.C.M. of denominators 6, 3 and 4 = 12, multiply each term with 12 to get:

$⇒ \dfrac{12(x + 2)}{6} - \dfrac{12(11 - x)}{3} + \dfrac{1 \times 12}{4} = \dfrac{12(3x - 4)}{12}$

⇒ 2(x + 2) - 4(11 - x) + 1 $\times$ 3 = 1(3x - 4)

⇒ (2x + 4) - (44 - 4x) + 3 = 3x - 4

⇒ 2x + 4 - 44 + 4x + 3 = 3x - 4

⇒ 6x - 37 = 3x - 4

⇒ 6x - 3x = 37 - 4

⇒ 3x = 33

⇒ x = $\dfrac{33}{3}$

⇒ x = 11

Hence, the value of x is 11.

Solve:

$\dfrac{2}{5x} - \dfrac{5}{3x} = \dfrac{1}{15}$

Answer

$\dfrac{2}{5x} - \dfrac{5}{3x} = \dfrac{1}{15}$

Since L.C.M. of denominators 5x and 3x = 15x, multiply each term with 15x to get:

$⇒ \dfrac{2 \times 15x}{5x} - \dfrac{5 \times 15x}{3x} = \dfrac{1 \times 15x}{15}$

⇒ 2 $\times$ 3 - 5 $\times$ 5 = 1 $\times$ x

⇒ 6 - 25 = x

⇒ - 19 = x

Hence, the value of x is -19.

Solve:

$\dfrac{x + 2}{3} - \dfrac{x + 1}{5} = \dfrac{x - 3}{4} - 1$

Answer

$\dfrac{x + 2}{3} - \dfrac{x + 1}{5} = \dfrac{x - 3}{4} - 1\\[1em] ⇒ \dfrac{x + 2}{3} - \dfrac{x + 1}{5} - \dfrac{x - 3}{4} = - 1$

Since L.C.M. of denominators 3, 5 and 4 = 60, multiply each term with 60 to get:

$⇒ \dfrac{60(x + 2)}{3} - \dfrac{60(x + 1)}{5} - \dfrac{60(x - 3)}{4} = - 1 \times 60$

⇒ 20(x + 2) - 12(x + 1) - 15(x - 3) = - 60

⇒ (20x + 40) - (12x + 12) - (15x - 45) = - 60

⇒ 20x + 40 - 12x - 12 - 15x + 45 = - 60

⇒ - 7x + 73 = - 60

⇒ - 7x = - 60 - 73

⇒ - 7x = - 133

⇒ x = $\dfrac{-133}{-7}$

⇒ x = 19

Hence, the value of x is 19.

Solve:

$\dfrac{3x - 2}{3} + \dfrac{2x + 3}{2} = x + \dfrac{7}{6}$

Answer

$\dfrac{3x - 2}{3} + \dfrac{2x + 3}{2} = x + \dfrac{7}{6}$

Since L.C.M. of denominators 3 and 2 = 6, multiply each term with 6 to get:

$⇒ \dfrac{6(3x - 2)}{3} + \dfrac{6(2x + 3)}{2} = x \times 6 + \dfrac{7 \times 6}{6}$

⇒ 2(3x - 2) + 3(2x + 3) = 6x + 7 $\times$ 1

⇒ (6x - 4) + (6x + 9) = 6x + 7

⇒ 6x - 4 + 6x + 9 = 6x + 7

⇒ 12x + 5 = 6x + 7

⇒ 12x - 6x = 7 - 5

⇒ 6x = 2

⇒ x = $\dfrac{2}{6}$

⇒ x = $\dfrac{1}{3}$

Hence, the value of x is $\dfrac{1}{3}$.

Solve:

$x -\dfrac{x - 1}{2} = 1 -\dfrac{x - 2}{3}$

Answer

$x -\dfrac{x - 1}{2} = 1 -\dfrac{x - 2}{3}\\[1em] ⇒ \dfrac{x}{1} -\dfrac{x - 1}{2} = 1 -\dfrac{x - 2}{3}\\[1em]$

Since L.C.M. of denominators 1, 2 and 3 = 6, multiply each term with 6 to get:

$⇒ \dfrac{x \times 6}{1} -\dfrac{6(x - 1)}{2} = 1 \times 6 -\dfrac{6(x - 2)}{3}$

⇒ x $\times$ 6 - 3(x - 1) = 6 - 2(x - 2)

⇒ 6x - (3x - 3) = 6 - (2x - 4)

⇒ 6x - 3x + 3 = 6 - 2x + 4

⇒ 3x + 3 = 10 - 2x

⇒ 3x + 2x = 10 - 3

⇒ 5x = 7

⇒ x = $\dfrac{7}{5}$

Hence, the value of x is $\dfrac{7}{5}$.

Solve:

$\dfrac{9x + 7}{2} - (x - \dfrac{x - 2}{7}) = 36$

Answer

$\dfrac{9x + 7}{2} - (x - \dfrac{x - 2}{7}) = 36\\[1em] ⇒ \dfrac{9x + 7}{2} - x + \dfrac{x - 2}{7} = 36\\[1em] ⇒ \dfrac{9x + 7}{2} - \dfrac{x}{1} + \dfrac{x - 2}{7} = 36$

Since L.C.M. of denominators 2 and 7 = 14, multiply each term with 14 to get:

$⇒ \dfrac{14(9x + 7)}{2} - \dfrac{x \times 14}{1} + \dfrac{14(x - 2)}{7} = 36 \times 14$