ICSE Class 7 Maths Chapter 16 Equations and Inequations

Chapter 16: Equations And Inequalities

Definition Of An Equation

An equation is a statement which states that the two expressions are equal.

For example:

(i) If the expressions 3x + 2 and x - 7 are equal, then 3x + 2 = x - 7 forms an equation.

(ii) The expressions 8y - 15 and \(\frac{y}{2}\) are equal \(\Rightarrow\) 8y - 15 = \(\frac{y}{2}\) is an equation.

To solve an equation means to find the value of the variable used in it.

For example, in equation x + 2 = 6 (where x is a variable)

\(\Rightarrow\) x = 6 - 2 \(\Rightarrow\) x = 4

Therefore x = 4 is the solution (root) of the equation x + 2 = 6

Important note: An equation remains unchanged if:

1. The same number is added to each side of the equation.

For example: x - 5 = 6 \(\Rightarrow\) x - 5 + 5 = 6 + 5 (Adding 5 on both the sides)

\(\Rightarrow\) x = 11

2. The same number is subtracted from each side of the equation.

For example: x + 5 = 6 \(\Rightarrow\) x + 5 - 5 = 6 - 5 (Subtracting 5 from both the sides)

\(\Rightarrow\) x = 1

3. The same number is multiplied to each side of the equation.

For example: \(\frac{x}{3}\) = 2 \(\Rightarrow\) \(\frac{x}{3}\) × 3 = 2 × 3 (Multiplying both the sides by 3)

\(\Rightarrow\) x = 6

4. Each side of the equation is divided by the same non-zero number.

For example: 4x = 12 \(\Rightarrow\) \(\frac{4x}{4}\) = \(\frac{12}{4}\) (Dividing both the sides by 4)

\(\Rightarrow\) x = 3

Example 1

Solve the equation x + 3 = 9.

Solution:

x + 3 = 9 \(\Rightarrow\) x + 3 - 3 = 9 - 3 (Subtracting 3 from both the sides)

\(\Rightarrow\) x = 6

Example 2

Solve:

(i) 2x + 5 = 11

(ii) \(\frac{x}{3}\) - 8 = 10

Solution:

(i) 2x + 5 = 11 \(\Rightarrow\) 2x + 5 - 5 = 11 - 5 (Subtracting 5 from both the sides)

\(\Rightarrow\) 2x = 6

\(\Rightarrow\) \(\frac{2x}{2}\) = \(\frac{6}{2}\) (Dividing each side by 2)

\(\Rightarrow\) x = 3

(ii) \(\frac{x}{3}\) - 8 = 10 \(\Rightarrow\) \(\frac{x}{3}\) - 8 + 8 = 10 + 8 (Adding 8 on both the sides)

\(\Rightarrow\) \(\frac{x}{3}\) = 18

\(\Rightarrow\) \(\frac{x}{3}\) × 3 = 18 × 3 (Multiplying both the sides by 3)

\(\Rightarrow\) x = 54

Short-Cut Method (Solving An Equation By Transposing Terms)

1. In an equation, if a + ve (positive) term is transposed (taken) from one side to the other, its sign is reversed, i.e., it becomes - ve (negative).

For example: x + 3 = 6 \(\Rightarrow\) x = 6 - 3 (Transposing + 3)

\(\Rightarrow\) x = 3

2. In an equation, if a - ve (negative) term is transposed from one side to the other, its sign becomes + ve (positive).

For example: x - 3 = 6 \(\Rightarrow\) x = 6 + 3 (Transposing - 3)

\(\Rightarrow\) x = 9

3. In an equation, if a term in multiplication is transposed to the other side, its sign is reversed, i.e., it goes in division.

For example: 3x = 6 \(\Rightarrow\) x = \(\frac{6}{3}\) (Transposing 3, which is in multiplication with x)

\(\Rightarrow\) x = 2

4. In an equation, if a term is in division, it is transposed to the other side in multiplication.

For example: \(\frac{x}{3}\) = 6 \(\Rightarrow\) x = 6 × 3 (Transposing 3, which is in division with x)

\(\Rightarrow\) x = 18

Although it is not a rule, the variable (x, y or z, etc.) is prefered to be kept on the left hand side of the equation.

Example 3

Solve:

(i) \(\frac{2}{3}\) x = 16

(ii) \(\frac{3}{4}\) x + 5 = 8

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

Solution:

(i) \(\frac{2}{3}\) x = 16 \(\Rightarrow\) 2x = 16 × 3 (Transposing 3)

\(\Rightarrow\) x = \(\frac{48}{2}\) (Transposing 2)

\(\Rightarrow\) x = 24

(ii) \(\frac{3}{4}\) x + 5 = 8 \(\Rightarrow\) \(\frac{3}{4}\) x = 8 - 5

\(\Rightarrow\) \(\frac{3}{4}\) x = 3 \(\Rightarrow\) 3x = 3 × 4

\(\Rightarrow\) x = \(\frac{12}{3}\) \(\Rightarrow\) x = 4

(iii) 5x - \(\frac{1}{2}\) = \(\frac{3}{4}\) \(\Rightarrow\) 5x = \(\frac{3}{4}\) + \(\frac{1}{2}\)

\(\Rightarrow\) 5x = \(\frac{5}{4}\) (since \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{3 + 2}{4}\) = \(\frac{5}{4}\))

\(\Rightarrow\) x = \(\frac{5}{4 \times 5}\) \(\Rightarrow\) x = \(\frac{1}{4}\)

Exercise 16 (A)

Solve the following equations:

1. x + 5 = 10

2. 2 + y = 7

3. a - 2 = 6

4. x - 5 = 8

5. 5 - d = 12

6. 3p = 12

7. 14 = 7m

8. 2x = 0

9. \(\frac{x}{9}\) = 2

10. \(\frac{y}{-12}\) = -4

11. 8x - 2 = 38

12. 2x + 5 = 5

13. 5x - 1 = 74

14. 14 = 27 - x

15. 10 + 6a = 40

16. c - \(\frac{1}{2}\) = \(\frac{1}{3}\)

17. \(\frac{a}{15}\) - 2 = 0

18. 12 = c - 2

19. 4 = x - 2.5

20. y + 5 = 8 \(\frac{1}{4}\)

21. x + \(\frac{1}{4}\) = -\(\frac{3}{8}\)

22. p + 0.02 = 0.08

23. p - 12 = 2\(\frac{2}{3}\)

24. -3x = 15

25. 1.3b = 39

26. \(\frac{5}{8}\) n = 20

27. \(\frac{3}{16}\) m = 21

28. 2a - 3 = 5

29. 3p - 1 = 8

30. 9y - 7 = 20

31. 2b - 14 = 8

32. \(\frac{7}{10}\) x + 6 = 41

33. \(\frac{5}{12}\) m - 12 = 48

Solving An Equation With The Variable On Both The Sides

Transpose the terms, containing the variable, to one side and the constants (i.e., terms without the variable) on the other side.

Example 4

Solve: 8x - 3 = 5x + 9

Solution:

8x - 5x = 9 + 3 (Transposing + 5x to left and -3 to the right side)

\(\Rightarrow\) 3x = 12

\(\Rightarrow\) x = \(\frac{12}{3}\) \(\Rightarrow\) x = 4

Example 5

Solve: 7 + 4x = 9x - 13

Solution:

7 + 4x = 9x - 13 OR 7 + 4x = 9x - 13

\(\Rightarrow\) 7 + 13 = 9x - 4x \(\Rightarrow\) 4x - 9x = -13 - 7

\(\Rightarrow\) 20 = 5x \(\Rightarrow\) -5x = -20

\(\Rightarrow\) \(\frac{20}{5}\) = x \(\Rightarrow\) x = \(\frac{-20}{-5}\)

\(\Rightarrow\) 4 = x x = 4

Example 6

Solve: 2(x - 5) + 3 (x - 2) = 8 + 7 (x - 4)

Solution:

2x - 10 + 3x - 6 = 8 + 7x - 28 (On removing the brackets)

\(\Rightarrow\) 5x - 16 = 7x - 20

\(\Rightarrow\) 5x - 7x = -20 + 16

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

Exercise 16 (B)

Solve:

1. 8y - 4y = 20

2. 9b - 4b + 3b = 16

3. 5y + 8 = 8y - 18

4. 6 = 7 + 2p - 5

5. 8 - 7x = 13x + 8

6. 4x - 5x + 2x = 28 + 3x

7. 9 + m = 6m + 8 - m

8. 24 = y + 2y + 3 + 4y

9. 19x + 13 - 12x + 3 = 23

10. 6b + 40 = -100 - b

11. 6 - 5m - 1 + 3m = 0

12. 0.4x - 1.2 = 0.3x + 0.6

13. 6(x + 4) = 36

14. 9(a + 5) + 2 = 11

15. 4(x - 2) = 12

16. -3(a - 6) = 24

17. 7(x - 2) = 2(2x - 4)

18. (x - 4) (2x + 3) = 2x²

19. 21 - 3(b - 7) = b + 20

20. x(x + 5) = x² + x + 32

Equations Involving Fractions

Example 7

Solve:

(i) \(\frac{x}{5}\) + x = 12

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

Solution:

(i) \(\frac{x}{5}\) + x = 12 \(\Rightarrow\) \(\frac{x}{5}\) + \(\frac{x}{1}\) = 12

\(\Rightarrow\) \(\frac{x + 5x}{5}\) = 12 (L.C.M. of 5 and 1 is 5)

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

\(\Rightarrow\) 6x = 12 × 5 \(\Rightarrow\) x = \(\frac{60}{6}\) \(\Rightarrow\) x = 10

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

\(\Rightarrow\) \(\frac{5(x - 1) - 3(2x - 3)}{15}\) = 1 (L.C.M. of 3 and 5 = 15)

\(\Rightarrow\) \(\frac{5x - 5 - 6x + 9}{15}\) = 1

\(\Rightarrow\) -x + 4 = 15 × 1

\(\Rightarrow\) -x = 15 - 4

\(\Rightarrow\) -x = 11 \(\Rightarrow\) x = -11

Example 8

Solve: y + 15% of y = 46

Solution:

y + \(\frac{15}{100}\) × y = 46 \(\Rightarrow\) y + \(\frac{3}{20}\) y = 46

\(\Rightarrow\) \(\frac{20y + 3y}{20}\) = 46

\(\Rightarrow\) \(\frac{23y}{20}\) = 46

\(\Rightarrow\) 23y = 46 × 20 \(\Rightarrow\) y = \(\frac{46 × 20}{23}\) \(\Rightarrow\) y = 40

Exercise 16 (C)

Solve:

1. \(\frac{x}{2}\) + x = 9

2. \(\frac{x}{5}\) + 2x = 33

3. \(\frac{3x}{4}\) + 4x = 38

4. \(\frac{x}{2}\) + \(\frac{x}{5}\) = 14

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

6. y + \(\frac{y}{2}\) = \(\frac{7}{4}\) - \(\frac{y}{4}\)

7. \(\frac{4x}{3}\) - \(\frac{7x}{3}\) = 1

8. \(\frac{1}{2}\) m + \(\frac{3}{4}\) m - m = 2.5

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

10. \(\frac{3a}{4}\) + \(\frac{a}{6}\) = 66

11. \(\frac{2p}{3}\) - \(\frac{p}{5}\) = 35

12. 0.6a + 0.2a = 0.4a + 8

13. p + 1.4p = 48

14. 10% of x = 20

15. y + 20% of y = 18

16. x - 30% of x = 35

17. \(\frac{x + 4}{2}\) + \(\frac{x}{3}\) = 7

18. \(\frac{y + 2}{3}\) + \(\frac{y + 5}{4}\) = 6

19. \(\frac{3a - 2}{7}\) - \(\frac{a - 2}{4}\) = 2

20. \(\frac{1}{2}\) (x + 5) - \(\frac{1}{3}\) (x - 2) = 4

21. \(\frac{x - 1}{2}\) - \(\frac{x - 2}{3}\) - \(\frac{x - 3}{4}\) = 0

22. \(\frac{x + 1}{3}\) + \(\frac{x + 4}{5}\) = \(\frac{x - 4}{7}\)

23. 15 - 2(5 - 3x) = 4(x - 3) + 13

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

25. 21 - 3(x - 7) = x + 20

26. \(\frac{3x - 2}{7}\) - \(\frac{x - 2}{4}\) = 2

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

28. \(\frac{x - 4}{7}\) = \(\frac{x + 3}{7}\) + \(\frac{x + 4}{5}\)

29. \(\frac{x}{5}\) - \(\frac{x}{3}\) = 1 - \(\frac{x - 2}{2}\)

30. 2x + 20% of x = 12.1

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