Simple Linear Equations — Exercise 14(E)

Solve for $x$:

$2x - 2 = \dfrac{1}{3}x - \dfrac{11}{4}$

Answer

Given,

⇒ $2x - 2 = \dfrac{1}{3}x - \dfrac{11}{4}$

L.C.M. of 3 and 4 = 12.

Multiplying both sides by 12,

$\Rightarrow (2x - 2) \times 12 = \left(\dfrac{1}{3}x - \dfrac{11}{4}\right) \times 12 \\[1em] \Rightarrow 2x \times 12 - 2 \times 12 = \dfrac{1}{3}x \times 12 - \dfrac{11}{4} \times 12 \\[1em] \Rightarrow 24x - 24 = 4x - 33 \\[1em] \Rightarrow 24x - 4x = -33 + 24 \\[1em] \Rightarrow 20x = -9 \\[1em] \Rightarrow x = -\dfrac{9}{20}.$

Hence, $\bm{x}$ = $-\dfrac{9}{20}$.

Solve for $x$:

$7(x - 4) - 2(x - 5) - 3(x - 2) = 3(2 - x) - 8$

Answer

Given,

⇒ $7(x - 4) - 2(x - 5) - 3(x - 2) = 3(2 - x) - 8$

$\Rightarrow 7x - 28 - 2x + 10 - 3x + 6 = 6 - 3x - 8 \\[1em] \Rightarrow 2x - 12 = -3x - 2 \\[1em] \Rightarrow 2x + 3x = -2 + 12 \\[1em] \Rightarrow 5x = 10 \\[1em] \Rightarrow x = 2.$

Hence, $\bm{x}$ = 2.

Solve for $x$:

$x - 40$% of $(x - 10)$ = 64

Answer

Solving,

$\Rightarrow x - 40$% of $(x − 10) = 64$

$\Rightarrow x - \dfrac{40}{100}(x - 10) = 64$

$\Rightarrow x - \dfrac{2}{5}(x - 10) = 64$

$\Rightarrow \dfrac{5x - 2(x - 10)}{5} = 64$

$\Rightarrow \dfrac{5x - 2x + 20}{5} = 64$

$\Rightarrow 3x + 20 = 320$

$\Rightarrow 3x = 300$

$\Rightarrow x = \dfrac{300}{3}$ = 100.

Hence, $\bm{x}$ = 100.

Solve for $x$:

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

Answer

Given,

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

L.C.M. of 3 and 2 = 6.

Multiplying both sides by 6,

$\Rightarrow \left(\dfrac{2-3x}{3} - \dfrac{3-5x}{2}\right) \times 6 = \dfrac{2}{3} \times 6 \\[1em] \Rightarrow 2(2 - 3x) - 3(3 - 5x) = 4 \\[1em] \Rightarrow 4 - 6x - 9 + 15x = 4 \\[1em] \Rightarrow 9x - 5 = 4 \\[1em] \Rightarrow 9x = 9 \\[1em] \Rightarrow x = 1.$

Hence, $\bm{x}$ = 1.

Five times a certain number diminished by 25 gives 25. Find the number.

Answer

Let the number be $x$.

Given,

Five times a certain number diminished by 25 gives 25.

$\therefore 5x - 25 = 25 \\[1em] \Rightarrow 5x = 25 + 25 \\[1em] \Rightarrow 5x = 50 \\[1em] \Rightarrow x = 10$

Hence, the number is 10.

Three-fifths of a number is greater than one-third of the same number by 4. Find the number.

Answer

Let the number be $x$.

Given,

Three-fifths of a number is greater than one-third of the same number by 4.

$\therefore \dfrac{3}{5}x - \dfrac{1}{3}x = 4 \\[1em] \Rightarrow \dfrac{9x - 5x}{15} = 4 \\[1em] \Rightarrow \dfrac{4x}{15} = 4 \\[1em] \Rightarrow 4x = 60 \\[1em] \Rightarrow x = 15$

Hence, the number is 15.

The sum of three consecutive even natural numbers is 36. Find the numbers.

Answer

Let the three consecutive even natural numbers be $x, x + 2$ and $x + 4$.

Given,

The sum of three consecutive even natural numbers is 36.

$\therefore x + (x + 2) + (x + 4) = 36 \\[1em] \Rightarrow 3x + 6 = 36 \\[1em] \Rightarrow 3x = 30 \\[1em] \Rightarrow x = 10$

∴ The numbers are 10, 10 + 2 = 12 and 10 + 4 = 14.

Hence, the numbers are 10, 12 and 14.

The sum of three consecutive odd integers is 51. Find the integers.

Answer

Let the three consecutive odd integers be $x, x + 2$ and $x + 4$.

Given,

The sum of three consecutive odd integers is 51.

$\therefore x + (x + 2) + (x + 4) = 51 \\[1em] \Rightarrow 3x + 6 = 51 \\[1em] \Rightarrow 3x = 45 \\[1em] \Rightarrow x = 15$

∴ The integers are 15, 15 + 2 = 17 and 15 + 4 = 19.

Hence, the integers are 15, 17 and 19.

The difference between the ages of a man and his son is 24 years. Two years hence, the age of the man will be three times the age of his son at that time. Find their present ages.

Answer

Let the present age of the son be $x$ years, then the present age of the man = ($x$ + 24) years.

Two years hence:

Age of son = ($x$ + 2) years

Age of man = ($x$ + 24 + 2) = ($x$ + 26) years

Given,

The difference between the ages of a man and his son is 24 years. Two years hence, the age of the man will be three times the age of his son at that time.

$\therefore x + 26 = 3(x + 2) \\[1em] \Rightarrow x + 26 = 3x + 6 \\[1em] \Rightarrow 26 - 6 = 3x - x \\[1em] \Rightarrow 20 = 2x \\[1em] \Rightarrow x = 10$

∴ Present age of man = $x$ + 24 = 10 + 24 = 34 years

Hence, the present age of the son is 10 years and that of the man is 34 years.

The difference between two supplementary angles is 20°. Find the angles.

Answer

Let the greater angle be $x$°, then the smaller angle = ($x$ − 20)°.

Since the angles are supplementary, their sum = 180°.

Given,

The difference between two supplementary angles is 20°.

$\therefore x + (x - 20) = 180 \\[1em] \Rightarrow 2x - 20 = 180 \\[1em] \Rightarrow 2x = 200 \\[1em] \Rightarrow x = 100$

∴ Smaller angle = $x$ − 20 = 100 − 20 = 80°

Hence, the angles are 100° and 80°.

Some articles are bought at ₹ 16 each and some other articles are bought at ₹ 5 each. If the total number of articles bought is 54 and their total cost is ₹ 600, find the number of articles of each kind.

Answer

Let the number of articles bought at ₹ 16 each be $x$, then the number of articles bought at ₹ 5 each = 54 − $x$.

Given,

Some articles are bought at ₹ 16 each and some other articles are bought at ₹ 5 each. If the total number of articles bought is 54 and their total cost is ₹ 600.

$\therefore 16x + 5(54 - x) = 600 \\[1em] \Rightarrow 16x + 270 - 5x = 600 \\[1em] \Rightarrow 11x = 600 - 270 \\[1em] \Rightarrow 11x = 330 \\[1em] \Rightarrow x = 30$

∴ Number of articles bought at ₹ 5 each = 54 − 30 = 24

Hence, the number of articles bought at ₹ 16 each is 30 and at ₹ 5 each is 24.

Find the lengths of the sides of an isosceles triangle whose each of the equal sides is 2 cm less than twice the length of the third side. Given perimeter of the isosceles triangle is 24 cm.

Answer

Let the third side of the isosceles triangle be $x$ cm.

Then, each of the equal sides = ($2x$ − 2) cm.

Perimeter of triangle = sum of its three sides

Given,

Perimeter = 24 cm

$\therefore (2x - 2) + (2x - 2) + x = 24 \\[1em] \Rightarrow 5x - 4 = 24 \\[1em] \Rightarrow 5x = 28 \\[1em] \Rightarrow x = \dfrac{28}{5} \text{ cm} \\[1em] \Rightarrow 2x - 2 = 2 \times \dfrac{28}{5} - 2 \\[1em] \Rightarrow 2x - 2 = \dfrac{56}{5} - 2 \\[1em] \Rightarrow 2x - 2 = \dfrac{56 - 10}{5} \\[1em] \Rightarrow 2x - 2 = \dfrac{46}{5} \text{ cm }$

Hence, the sides of the triangle are $\dfrac{46}{5} \text{ cm, } \dfrac{46}{5} \text{ cm and } \dfrac{28}{5}$ cm.

Divide 600 into two parts such that one-fourths of one part exceeds one-sixths of the other part by 20.

Answer

Let one part be $x$, then the other part = 600 − $x$.

Given,

Divide 600 into two parts such that one-fourths of one part exceeds one-sixths of the other part by 20.

$\therefore \dfrac{1}{4}x - \dfrac{1}{6}(600 - x) = 20 \\[1em] \Rightarrow \dfrac{3x - 2(600 - x)}{12} = 20 \\[1em] \Rightarrow 3x - 1200 + 2x = 240 \\[1em] \Rightarrow 5x = 1440 \\[1em] \Rightarrow x = 288$

∴ Other part = 600 − 288 = 312

Hence, the two parts are 288 and 312.

Solve for $x$:

$\dfrac{3}{4} (2x - 5) - \dfrac{5}{6} (7 - 5x) = \dfrac{7x}{3}$

Answer

Given,

⇒ $\dfrac{3}{4} (2x - 5) - \dfrac{5}{6} (7 - 5x) = \dfrac{7x}{3}$

L.C.M. of 4, 6 and 3 = 12.

Multiplying both sides by 12,

$\Rightarrow \dfrac{3}{4}(2x - 5) \times 12 - \dfrac{5}{6}(7 - 5x) \times 12 = \dfrac{7x}{3} \times 12 \\[1em] \Rightarrow 9(2x - 5) - 10(7 - 5x) = 28x \\[1em] \Rightarrow 18x - 45 - 70 + 50x = 28x \\[1em] \Rightarrow 68x - 115 = 28x \\[1em] \Rightarrow 68x - 28x = 115 \\[1em] \Rightarrow 40x = 115 \\[1em] \Rightarrow x = \dfrac{115}{40} \\[1em] \Rightarrow x = \dfrac{23}{8}.$

Hence, $\bm{x}$ = $\dfrac{23}{8}$.

Solve for $x$:

$\dfrac{3}{4} (7x - 1) - \left(2x - \dfrac{1-x}{2}\right) = x + \dfrac{3}{2}$

Answer

Given,

⇒ $\dfrac{3}{4} (7x - 1) - \left(2x - \dfrac{1-x}{2}\right) = x + \dfrac{3}{2}$

Multiplying both sides by 4,

$\Rightarrow \dfrac{3}{4}(7x - 1)\times 4 - \left(2x - \dfrac{1 - x}{2}\right)\times 4 = \Big(x + \dfrac{3}{2}\Big) \times 4 \\[1em] \Rightarrow 3(7x - 1) - \left(8x - 2(1 - x)\right) = 4x + 6 \\[1em] \Rightarrow 21x - 3 - (8x - 2 + 2x) = 4x + 6 \\[1em] \Rightarrow 21x - 3 - (10x - 2) = 4x + 6 \\[1em] \Rightarrow 21x - 3 - 10x + 2 = 4x + 6 \\[1em] \Rightarrow 11x - 1 = 4x + 6 \\[1em] \Rightarrow 11x - 4x = 6 + 1 \\[1em] \Rightarrow 7x = 7 \\[1em] \Rightarrow x = 1.$

Hence, $\bm{x}$ = 1.