ICSE Class 8 Maths Chapter 18 Linear Equations in One Variable

Chapter 18: Linear Equations In One Variable

With problems based on linear equations

18.1 Review

1. To solve an equation means to find the value of its variable (i.e. x, y or z, etc).
2. A linear equation has only one solution, which is called its root.

An equation remains unaltered (unchanged) on:
(i) adding the same number to both sides of it;
(ii) subtracting the same number from both sides of it;
(iii) multiplying both sides of it by the same number, and
(iv) dividing both sides of it by the same number.

Test Yourself

1. x + 3 = 5 \(\Rightarrow\) x + 3 - 3 = ......... \(\Rightarrow\) x = .........
2. y - 2 = 4 \(\Rightarrow\) y - 2 + 2 = ......... \(\Rightarrow\) y = .........
3. 3z = 15 \(\Rightarrow\) \(\frac{3z}{3}\) = ......... \(\Rightarrow\) z = .........
4. \(\frac{x}{5}\) = -2 \(\Rightarrow\) \(\frac{x}{5}\) × 5 = ......... \(\Rightarrow\) x = .........
5. 3y - 2 = 10 \(\Rightarrow\) 3y = ......... \(\Rightarrow\) y = .........
6. \(\frac{m}{5}\) + 7 = 9 \(\Rightarrow\) \(\frac{m}{5}\) = ................. \(\Rightarrow\) m = .................
7. 4x - 13 = 7 - x \(\Rightarrow\) 4x + x = ................. \(\Rightarrow\) 5x = ......... and x = .................
8. 2(7z - 3) = 3(4z - 2) \(\Rightarrow\) ................. = ...............
\(\Rightarrow\) 14z - 12z = ................. \(\Rightarrow\) 2z = ......... and z = .........

Example 1

Solve: 21 - 3(a - 7) = a + 20

Solution

21 - 3a + 21 = a + 20
\(\Rightarrow\) 42 - 20 = a + 3a \(\Rightarrow\) 22 = 4a \(\therefore\) a = \(\frac{22}{4}\) = 5\(\frac{1}{2}\)
(Ans.)

Example 2

Solve: \(\frac{y+2}{4}\) - \(\frac{y-3}{3}\) = \(\frac{1}{2}\)

Solution

Since, L.C.M. of denominators 4, 3 and 2 = 12
\(\therefore\) 12 × \(\frac{y+2}{4}\) - 12 × \(\frac{y-3}{3}\) = 12 × \(\frac{1}{2}\) [Multiplying each term by 12]
\(\Rightarrow\) 3(y + 2) - 4(y - 3) = 6
\(\Rightarrow\) 3y + 6 - 4y + 12 = 6
\(\Rightarrow\) -y = -12 \(\Rightarrow\) y = 12 (Ans.)

Example 3

Solve: (i) \(\frac{5}{x}\) = \(\frac{7}{x-4}\) (ii) \(\frac{a-2}{a+4}\) = \(\frac{a-3}{a+1}\)

Solution

On cross-multiplying; we get:
(i) 7x = 5(x - 4)
\(\Rightarrow\) 7x = 5x - 20
\(\Rightarrow\) 7x - 5x = -20
\(\Rightarrow\) 2x = -20
\(\Rightarrow\) x = -10 (Ans.)

(ii) (a - 2)(a + 1) = (a - 3)(a + 4)
\(\Rightarrow\) a² - 2a + a - 2 = a² - 3a + 4a - 12
\(\Rightarrow\) a² - a - 2 = a² + a - 12
\(\Rightarrow\) a² - a - a² - a = -12 + 2
\(\Rightarrow\) -2a = -10
\(\Rightarrow\) a = 5 (Ans.)

Example 4

Solve: \(\frac{2x+1}{10}\) - \(\frac{3-2x}{15}\) = \(\frac{x-2}{6}\). Hence, find the value of y, if \(\frac{1}{x}\) + \(\frac{1}{y}\) = 3.

Solution

Since, L.C.M. of denominators 10, 15 and 6 = 30, multiply each fraction by 30 to get:
30 × \(\frac{2x+1}{10}\) - 30 × \(\frac{3-2x}{15}\) = 30 × \(\frac{x-2}{6}\)
\(\Rightarrow\) 3(2x + 1) - 2(3 - 2x) = 5(x - 2)
\(\Rightarrow\) 6x + 3 - 6 + 4x = 5x - 10
\(\Rightarrow\) 10x - 5x = -10 + 3
\(\Rightarrow\) 5x = -7 and x = -\(\frac{7}{5}\) (Ans.)

Now, \(\frac{1}{x}\) + \(\frac{1}{y}\) = 3 \(\Rightarrow\) -\(\frac{5}{7}\) + \(\frac{1}{y}\) = 3 [x = -\(\frac{7}{5}\) \(\Rightarrow\) \(\frac{1}{x}\) = -\(\frac{5}{7}\)]
\(\Rightarrow\) \(\frac{1}{y}\) = 3 + \(\frac{5}{7}\) = \(\frac{26}{7}\)

\(\therefore\) y = \(\frac{7}{26}\) (Ans.)

Exercise 18 (A)

Solve the following equations:
1. 20 = 6 + 2x
2. 15 + x = 5x + 3
3. \(\frac{3x+2}{x-6}\) = -7
4. 3a - 4 = 2(4 - a)
5. 3(b - 4) = 2(4 - b)
6. \(\frac{x+2}{9}\) = \(\frac{x+4}{11}\)
7. \(\frac{x-8}{5}\) = \(\frac{x-12}{9}\)
8. 5(8x + 3) = 9(4x + 7)
9. 3(x + 1) = 12 + 4(x - 1)
10. \(\frac{3x}{4}\) - \(\frac{1}{4}\)(x - 20) = \(\frac{x}{4}\) + 32
11. 3a - \(\frac{1}{5}\) = \(\frac{a}{5}\) + 5\(\frac{2}{5}\)
12. \(\frac{x}{3}\) - 2\(\frac{1}{2}\) = \(\frac{4x}{9}\) - \(\frac{2x}{3}\)
13. \(\frac{4(y+2)}{5}\) = 7 + \(\frac{5y}{13}\)
14. \(\frac{a+5}{6}\) - \(\frac{a+1}{9}\) = \(\frac{a+3}{4}\)
15. \(\frac{2x-13}{5}\) - \(\frac{x-3}{11}\) = \(\frac{x-9}{5}\) + 1
16. 6(6x - 5) - 5(7x - 8) = 12(4 - x) + 1
17. (x - 5)(x + 3) = (x - 7)(x + 4)
18. (x - 5)² - (x + 2)² = -2
19. (x - 1)(x + 6) - (x - 2)(x - 3) = 3
20. \(\frac{3x}{x+6}\) - \(\frac{x}{x+5}\) = 2
21. \(\frac{1}{x-1}\) + \(\frac{2}{x-2}\) = \(\frac{3}{x-3}\)
22. \(\frac{x-1}{7x-14}\) = \(\frac{x-3}{7x-26}\)
23. \(\frac{1}{x-1}\) - \(\frac{1}{x}\) = \(\frac{1}{x+3}\) - \(\frac{1}{x+4}\)
24. Solve: \(\frac{2x}{3}\) - \(\frac{x-1}{6}\) + \(\frac{7x-1}{4}\) = 2\(\frac{1}{6}\). Hence, find the value of 'a', if \(\frac{1}{a}\) + 5x = 8.
25. Solve: \(\frac{4-3x}{5}\) + \(\frac{7-x}{3}\) + 4\(\frac{1}{3}\) = 0. Hence, find the value of 'p', if 3p - 2x + 1 = 0.

Teacher's Note

Linear equations are used daily when calculating distances, speeds, and time in travel planning. For instance, when determining how long a journey will take at a constant speed, you are solving a linear equation.

18.2 To Solve Problems Based On Linear Equations

Steps

1. Read the problem carefully to know what is given and what is to be found.
2. Represent the unknown quantity as x or some other letter such as a, b, y, z, etc.
3. According to the conditions, given in the problem, write the relation between the given quantity and quantity to be found.
4. Solve the equation to obtain the value of the unknown.

Example 5

Find a number such that one-fifth of it is less than its one-fourth by 3.

Solution

Let the required number be x.

Since, one-fifth of x = \(\frac{x}{5}\) and one-fourth of it = \(\frac{x}{4}\); then according to the given statement:

\(\frac{x}{4}\) - \(\frac{x}{5}\) = 3 \(\Rightarrow\) \(\frac{5x-4x}{20}\) = 3 [L.C.M. of 4 and 5 = 20]
\(\Rightarrow\) x = 3 × 20 = 60 (Ans.)

Example 6

The difference of the squares of two consecutive even natural numbers is 92. Taking x as the smaller of the two numbers, form an equation in x and hence find the larger of the two.

Solution

Since, the consecutive even natural numbers differ by 2 and it is given that the smaller of the two numbers is x; therefore, the next (larger) even number is x + 2.

According to the given statement:

(x + 2)² - x² = 92 [Difference of the squares]

\(\Rightarrow\) x² + 4x + 4 - x² = 92
\(\Rightarrow\) 4x = 92 - 4 = 88
\(\Rightarrow\) x = 22

\(\therefore\) Larger even number = x + 2 = 22 + 2 = 24 (Ans.)

In case of integers, natural numbers and whole numbers:
1. Consecutive numbers are taken as: x + 1, x + 2, ...........
2. Consecutive even numbers are taken as: x, x + 2, x + 4, ...........; where x is an ...........
3. Consecutive odd numbers are also taken as: x, x + 2, x + 4, ...........; where x is an ...........
4. Consecutive multiples of 3 are taken as: x, x + 3, x + 6, ...........; ...........

Example 7

A rectangle is 8 cm long and 5 cm wide. Its perimeter is doubled when each of its sides is increased by x cm. Form an equation in x and find the new length of the rectangle.

Solution

Since, length of the rectangle = 8 cm and its width = 5 cm
Its perimeter = 2(length + width)
= 2(8 + 5)cm = 26 cm

On increasing each of its sides by x cm,
its new length = (8 + x) cm
and, new width = (5 + x) cm

Its new perimeter = 2(8 + x + 5 + x) cm
= (26 + 4x) cm

Given: new perimeter = 2 times the original perimeter
\(\Rightarrow\) 26 + 4x = 2 × 26
\(\Rightarrow\) 4x = 52 - 26 = 26
\(\Rightarrow\) x = \(\frac{26}{4}\) = 6.5 cm

And, the new length of the rectangle = (8 + x) cm
= (8 + 6.5) cm = 14.5 cm. (Ans.)

Example 8

A man is 24 years older than his son. In 2 years, his age will be twice the age of his son. Find their present ages.

Solution

Let the present age of the son be x years
\(\therefore\) Present age of the father = (x + 24) years

In 2 years:
The man's age will be (x + 24) + 2 = (x + 26) years
and son's age will be x + 2 years

According to the problem: x + 26 = 2(x + 2)
On solving we get: x = 22

\(\therefore\) Present age of the man = x + 24 = 22 + 24 = 46 years
and, Present age of the son = x = 22 years. (Ans.)

Teacher's Note

Age-related problems are common in real life, such as determining when a parent's age will be a certain multiple of a child's age, which helps in understanding family relationships mathematically.

Example 9

One day a boy walked from his house to his school at the speed of 4 km/hr and he reached ten minutes late to the school. Next day, he ran at the speed of 8 km/hr and was 5 minutes early to the school. Find the distance between his house and school.

Solution

Let the distance between his house and school be x km.

Since, time = \(\frac{\text{distance}}{\text{speed}}\)

\(\therefore\) To reach the school, first day he takes \(\frac{x}{4}\) hrs and next day he takes \(\frac{x}{8}\) hrs.

Since, the difference of two timings = 10 minutes + 5 minutes = 15 minutes = \(\frac{1}{4}\) hrs

\(\therefore\) \(\frac{x}{4}\) - \(\frac{x}{8}\) = \(\frac{1}{4}\)

On solving, we get: x = 2 km. (Ans.)

Example 10

Two consecutive even numbers are such that half of the larger exceeds one-fourth of the smaller by 5. Find the numbers.

Solution

Let the required even numbers be x and x + 2.

Given: \(\frac{1}{2}\)(x + 2) - \(\frac{1}{4}\) x = 5

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

\(\Rightarrow\) x + 4 = 20 i.e. x = 20 - 4 = 16

\(\therefore\) Required numbers = x and x + 2
= 16 and 16 + 2 = 16 and 18 (Ans.)

Example 11

A person is paid ₹ 150 for each day he works and is fined ₹ 30 for each day he remains absent. If in 40 days, he earned ₹ 3,300; find for how many days did he work?

Solution

Let the man works for x days
\(\therefore\) He remains absent for (40 - x) days.

Since, the man gets ₹ 150 for each day he worked and is fined ₹ 30 for each day he remains absent.

\(\therefore\) 150x - 30 (40 - x) = 3,300
\(\Rightarrow\) 150x - 1200 + 30x = 3,300
\(\Rightarrow\) 180x = 3,300 + 1,200 = 4,500

and, x = \(\frac{4,500}{180}\) = 25

\(\therefore\) The man worked for 25 days (Ans.)

Test Yourself

9. Two numbers differ by 5; if one number is x; the other number = .............. or ..............
10. (i) Consecutive multiples of 8, differ by ..............
(ii) Consecutive multiple of 15, differ by ..............
11. If x years and (2x - 7) years are the present ages of A and B respectively, their ages:
(a) 6 years ago were, A = .............. ........ and B = ................................... = ..............
(b) 10 years hence, will be, A = .............. ........ and B = ................................... = ..............

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