ICSE Class 8 Maths Algebra Chapter 19 Simple Linear Equations

Simple Linear Equations

Simple Linear Equation

Root

Transposition

Linear Equations

An equation states that the algebraic expression on the left-hand side (LHS) of the equality sign is equal to the algebraic expression on the right-hand side (RHS).

A simple linear equation is an algebraic statement that involves only one variable, the degree of which is not more than 1. Solving an equation involves finding the root of the equation or the value of the variable for which the equation is true.

Solving an equation generally involves grouping the variable terms and the constant terms on opposite sides of the equation by transposition.

On transposition of a term to its opposite side,

its sign changes: \(x + 17 = 23 \Rightarrow x = 23 - 17\)

a multiplier changes into a divisor: \(18x = 126 \Rightarrow x = \frac{126}{18}\)

a divisor changes into a multiplier: \(\frac{x}{3} = 28 \Rightarrow 14x = 28 \times 3\)

Example 1: Solve \(5(x - 2) + 2(x + 4) = 9(x - 4)\)

\(\Rightarrow 5x - 10 + 2x + 8 = 9x - 36\)

\(\Rightarrow 7x - 2 = 9x - 36\)

\(\Rightarrow 36 - 2 = 9x - 7x\)

(transposing variables and constant terms)

\(\Rightarrow 2x = 34\)

\(\Rightarrow x = \frac{34}{2} = 17\)

CHECK: \(5(17 - 2) + 2(17 + 4) = 9(17 - 4)\)

\(\Rightarrow 5 \times 15 + 2 \times 21 = 9 \times 13\)

\(\Rightarrow 75 + 42 = 117\)

\(\Rightarrow 117 = 117\)

Example 2: Solve \(\frac{x + 8}{7} + \frac{x - 1}{3} = \frac{x}{2}\)

\(\Rightarrow \frac{3(x + 8) + 7(x - 1)}{21} = \frac{x}{2}\)

\(\Rightarrow \frac{3x + 24 + 7x - 7}{21} = \frac{x}{2}\)

\(\Rightarrow \frac{10x + 17}{21} = \frac{x}{2}\)

\(\Rightarrow 2(10x + 17) = 21x\)

(transposing denominators, or cross-multiplying)

\(\Rightarrow 20x + 34 = 21x\)

\(\Rightarrow 34 = 21x - 20x\)

\(\Rightarrow x = 34\)

Example 3: Solve \(\frac{4(2 - x) - 3(x + 5)}{4 - x} = 14\)

\(\Rightarrow 8 - 4x - 3x - 15 = 14(4 - x)\)

\(\Rightarrow -7x - 7 = 56 - 14x\)

\(\Rightarrow 14x - 7x = 56 + 7\)

\(\Rightarrow 7x = 63\)

\(\Rightarrow x = \frac{63}{7} = 9\)

Teacher's Note

Linear equations are the foundation for solving real-world problems, from calculating discounts while shopping to determining the right mix of ingredients in cooking.

Example 4: Solve

\(2x - 2 - \frac{1}{3}\left[3x - 3 - \frac{1}{4}(7x + 7 - 5x - 5)\right] = 7\)

\(\Rightarrow 2x - 2 - \frac{1}{3}\left[3x - 3 - \frac{1}{4}(2x + 12)\right] = 7\)

\(\Rightarrow 2x - 2 - \frac{1}{3}\left[3x - 3 - \frac{x}{2} - 3\right] = 7\)

\(\Rightarrow 2x - 2 - \frac{1}{3}\left[\frac{6x - x}{2} - 6\right] = 7\)

\(\Rightarrow 2x - 2 - \frac{1}{3}\left[\frac{5x}{2} - 6\right] = 7\)

\(\Rightarrow 2x - 2 - \frac{5x}{6} + 2 = 7\)

\(\Rightarrow \frac{12x - 5x}{6} = 7\)

\(\Rightarrow 7x = 42\)

\(\Rightarrow x = \frac{42}{7} = 6\)

Try this!

1. Solve \(9a - 94 = 100\)

2. Solve \(\frac{x + 7}{x + 5} = 17\)

Exercise 19.1

Solve the following equations.

1. \(7x + 35 = 84\)

2. \(14x - 32 = 80\)

3. \(11x - 14 = 8x + 22\)

4. \(13x + 23 = 16x - 16\)

5. \(6(x - 6) = 4x\)

6. \(5(x + 7) = 7.5x\)

7. \(8(x + 9) = 20(x - 3)\)

8. \(11(x - 5) = 6(x + 5)\)

9. \(4(x - 3) + 2(x + 5) = 8(x - 4)\)

10. \(5(x + 3) - 2(x - 2) = 5(x - 5)\)

11. \(5(x + 7) - 2(x - 3) = 7(x - 1)\)

12. \(3(x - 2) + 2(x - 5) = \frac{9x}{2}\)

13. \(\frac{x + 8}{x - 4} = 2\)

14. \(\frac{x + 18}{x - 18} = 5\)

15. \(\frac{x - 2}{7} + \frac{x + 2}{3} = \frac{x}{2}\)

16. \(\frac{x - 3}{5} + \frac{x + 2}{6} = \frac{5x}{14}\)

17. \(\frac{x}{4} - \frac{x}{7} = \frac{x - 2}{9}\)

18. \(\frac{x}{2} - \frac{x}{3} = \frac{x - 4}{5}\)

19. \(\frac{5x + 3}{3} - \frac{3x + 4}{4} = \frac{2x + 9}{3} = 0\)

20. \(0.75x - 0.5x + 3 = 0.3x\)

21. \(0.4x + 1.7x = 2.15x - 1.1\)

22. \(\frac{x}{3} + \frac{x}{4} - \frac{x}{5} + 13\frac{4}{5} = \)

23. \(\frac{x}{6} - \frac{x}{7} - \frac{x}{8} - 4\frac{1}{4} = \)

24. \(\frac{3x - 5(1 - x)}{x - 4} = 5\)

25. \(\frac{3(3x - 5) - 3(x - 3)}{x + 3} = 4\)

26. \(\frac{x - (4 - 6x)}{9x - (x - 4)} = \frac{2}{3}\)

27. \(\frac{4(3 - x) - 2(x + 10)}{2 - x} = 10\)

Challenge

1. \(\frac{2x + 6}{11} - \frac{3x - 6}{6} - \frac{4x - 7}{5} = 0\)

2. \(x + 12 + [3x + 1 + (9x - 5 - (5x + 3 - 2x - 5))] = 30\)

3. \(\frac{1}{5}\left[2x + 2 - \frac{1}{7}(6x + 6 - 3x + 2)\right] = 5\)

Solving Word Problems With Equations

Example 5: Rajat is 4 years elder to his wife. If one-half of his age exceeds one-third of his wife's age by 9 years, find how old Rajat and his wife are.

Let Rajat's wife's age = x

Then Rajat's age = x + 4

Given \(\frac{1}{2}(x + 4) - \frac{x}{3} = 9\)

\(\Rightarrow \frac{x + 4}{2} - \frac{x}{3} = 9\)

\(\Rightarrow \frac{3x + 12 - 2x}{6} = 9\)

\(\Rightarrow x + 12 = 54\)

\(\Rightarrow x = 54 - 12 = 42\)

Thus, Rajat's wife is 42 years old and Rajat is 42 + 4 = 46 years old.

Example 6: Divide Rs 7100 between A and B such that one-fifth of A's share exceeds one-sixth of B's share by Rs 155.

If A's share = x, then B's share = 7100 - x

Now \(\frac{x}{5} - \frac{1}{6}(7100 - x) = 155\)

\(\Rightarrow \frac{x}{5} - \frac{7100 - x}{6} = 155\)

\(\Rightarrow \frac{6x - 35500 + 5x}{30} = 155\)

\(\Rightarrow 11x - 35500 = 155 \times 30\)

\(\Rightarrow 11x = 4650 + 35500\)

\(\Rightarrow 11x = 40150\)

\(\Rightarrow x = \frac{40150}{11} = 3650\)

Thus, A's share = Rs 3650 and B's share = Rs 7100 - 3650 = Rs 3450

Example 7: If the length of a square is decreased by 7 cm, its area decreases by 259 cm². What was the original length of the square?

Let the original length of the square be x.

Then the original area of the square = x²

New length of the square = x - 7

New area of the square = (x - 7)² = x² - 14x + 49

Given that original area - new area = 259 cm²

\(\Rightarrow x² - (x² - 14x + 49) = 259\)

\(\Rightarrow x² - x² + 14x - 49 = 259\)

\(\Rightarrow 14x = 259 + 49\)

\(\Rightarrow x = \frac{308}{14} = 22\)

Thus, the original length of the square was 22 cm.

If the face values of the two digits in a two-digit number are x and y, the number can be either 10x + y or 10y + x. For instance, 3 × 10 + 9 = 39 or 9 × 10 + 3 = 93

Example 8: The sum of the two digits in a 2-digit number is 13. If 45 is added to the number, the digits are reversed. Find the original 2-digit number.

Now let the digit in the ones place of the 2-digit number in this example be x.

Then the digit in the tens place = 13 - x and the number = 10 (13 - x) + x

When the number is reversed, the digit in the ones place = 13 - x, the digit in the tens place = x and the new number = 10x + (13 - x)

Given \({10x + (13 - x)} - {10(13 - x) + x} = 45\)

\(\Rightarrow (10x + 13 - x) - (130 - 10x + x) = 45\)

\(\Rightarrow (9x + 13) - (130 - 9x) = 45\)

\(\Rightarrow 9x + 13 - 130 + 9x = 45\)

\(\Rightarrow 18x - 117 = 45\)

\(\Rightarrow x = \frac{45 + 117}{18} = \frac{162}{18} = 9\)

Thus, the digit in the ones place of the original number is 9 and the digit in the tens place = 13 - 9 = 4 or the original number = 49

CHECK: 49 + 45 = 94 (digits reversed)

Example 9: Mihir starts from a jetty and swims against the current of the Hooghly for 17 \(\frac{1}{2}\) minutes to reach the temple ghat. On the return trip, he

Teacher's Note

Word problems teach us to translate real-world situations into mathematical language, a skill useful in budgeting, planning, and decision-making in everyday life.

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