WBBSE Solutions For Class 8 Maths Algebra Chapter 8 Formation Of An Equation And Its Solutions

Algebra Chapter 8 Formation Of An Equation And Its Solutions

Equation: An equation is a statement of equality of two algebraic expressions, which involve one or more unknown quantities called the “variables”.

Example:

1. 3x + 5 = 0
2. 3x – 5 = x + 7
3. y + \(\frac{1}{5}\) = \(\frac{y}{7}\) − 3
4. x² + 9 = 25 are equations.

⇒ In the above example equations (1), (2) and (3) involve only linear polynomials where as (4) is quadratic polynomial.

⇒ Linear equation: An equation involving linear polynomials is called a linear equation.

⇒ Solution of a linear equation: A value of the variable which when substituted for the variable in the equation makes the two sides of the equation equal is called the solution of the equation.

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Consider the equation 4x – 3 = 5.

⇒ If we substitute the value 2 for x, we get,

LHS = 4 x 2 – 3 = 5 = RHS

∴ x = 2 is a solution of the equation 4x – 3 = 5.

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⇒ If we substitute the value 3 for x, we get, LHS = 4 x 3 – 3= 9 ≠ RHS.

⇒ So x = 3 is not a solution of equation 4x – 3 = 5

Solving an equation: Solving a linear equation means finding a value of the variable which satisfies the equation.

Algebra Chapter 8 Formation Of An Equation And Its Solutions Examples

Example 1. In the following equations, verify whether the given value of the variable is a solution of the equation.

1. \(\frac{x}{2}\) + \(\frac{x}{3}\) = 2; x = 6
2. 2(x – 3) + 5(x + 6)= 3; x = -3

Solution:

1. LHS = \(\frac{x}{2}\) + \(\frac{x}{3}\) = \(\frac{6}{2}\) + \(\frac{6}{3}\) = 3 + 2 = 5 ≠ RHS

So x = 6 is not the solution of given equation.

2. LHS, 2(x – 3) + 5(x + 6)

= 2(-3 – 3) + 5(-3 + 6) = 2x – 6 + 5 x 3 = 12 + 15 = 3 = RHS

So x= -3 is the solution of the given equation.

To solve an equation, we use the following proporties of equality:

1. Some quantity can be added to both sides of an equation without changing the equality.
2. Some quantity can be subtracted from both the sides of an equation without changing the equality.
3. Both the sides of an equation can be multiplied by the same non-zero number without changing the equality.
4. Both the sides of an equation may be divided by the same non-zero number without changing the equality.

Illustrative examples:

Solve the following equation:

1. 12x – (3x – 5) – [7 – {2x -(2x – 5)}] = 25

Solution

Given

⇒ 12x – 3x + 5 – [7 – {2x – 2x + 5}] = 25

⇒ 9x + 5 -(7 – 5) = 25

⇒ 9x + 5 – 2 = 25

⇒ 9x + 3 = 25

⇒ 9x = 25 – 3

⇒ 9x = 22

⇒ x = \(\frac{22}{9}\)

2. 2(x – 2) (x + 4) + 3(x² + 4) = 5(x² – 5x + 6)

Solution.

Given

⇒ 2(x² + 4x – 2x – 8) + 3x² + 12

⇒ 5x² – 25x + 30

⇒ 2x² + 8x – 4x – 16 + 3x² + 12 = 5x² – 25x + 30

⇒ 5x² + 4x – 4 = 5x – 25x + 30

⇒ 5x² + 4x – 5x² + 25x = 30 + 4

⇒ 29x = 34

⇒ x = \(\frac{34}{29}\)

3. \(\frac{3}{x-1}+\frac{1}{x+1}=\frac{4}{x}[x \neq 0,1,-1]\)

Solution

Given

⇒ \(\frac{3}{x-1}+\frac{1}{x+1}=\frac{1}{x}+\frac{3}{x}\)[because 4=1+3]

⇒ \(\frac{3}{x-1}-\frac{3}{x}=\frac{1}{x}-\frac{1}{x+1}\)

⇒ \(\frac{3 x-3 x+3}{x(x-1)}=\frac{x+1-x}{x(x+1)}\)

⇒ \(\frac{3}{x(x-1)}=\frac{1}{x(x+1)}\)

⇒ \(\frac{3}{x-1}=\frac{1}{x+1}\) [x ≠ 0, multilying both sides by x]

⇒ 3(x + 1) = x – 1

⇒ 3x + 3 = x – 1

⇒ 3x – x = – 1 – 3

⇒ 2x = -4

⇒ x = –\(\frac{4}{x}\)

⇒ x = -2

4. \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x-3 a-3 b}{a+b}=0\)

Solution.

Given

⇒ \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x-3(a+b)}{a+b}=0\)

⇒ \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x}{a+b}-3=0\)

⇒ \(\left(\frac{x-a}{b}-1\right)+\left(\frac{x-b}{a}-1\right)+\left(\frac{x}{a+b}-1\right)=0\)

⇒ \(\frac{x-a-b}{b}+\frac{x-b-a}{a}+\frac{x-a-b}{a+b}=0\)

⇒ \((x-a-b)\left(\frac{1}{b}+\frac{1}{a}+\frac{1}{a+b}\right)=0\)

⇒ \(x-a-b=0\left[because \frac{1}{a}+\frac{1}{a}+\frac{1}{a+b} \neq 0\right]\)

⇒ x = a + b

5. \(\frac{3 y+1}{16}+\frac{2 y-3}{7}=\frac{y+3}{8}+\frac{3 y-1}{14}\)

Solution.

Given

⇒ \(\frac{3 y+1}{16}-\frac{y+3}{8}=\frac{3 y-1}{14}-\frac{2 y-3}{7}\)

⇒ \(\frac{3 y+1-2 y-6}{16}=\frac{3 y-1-4 y+6}{14}\)

⇒ \(\frac{y-5}{16}=\frac{5-y}{14}\)

⇒ \(\frac{y-5}{8}=\frac{5-y}{7}\) [Multiplying both side by 2]

⇒ 7(y – 5) = 8(5 – y)

⇒ 7y – 35 = 40 – 8y

⇒ 7y + 8y = 40 + 35

⇒ 15y = 75

⇒ y = \(\frac{75}{15}\)

⇒ y = 5

6. \(\frac{9 x+5}{14}+\frac{8 x-7}{7}=\frac{18 x+11}{28}+\frac{5}{4}\)

Solution.

Given

⇒ \(\frac{9 x+5+16 x-14}{14}=\frac{18 x+11}{28}+\frac{5}{4}\)

⇒ \(\frac{25 x-9}{14}-\frac{18 x+11}{28}=\frac{5}{4}\)

⇒ \(\frac{50 x-18-18 x-11}{28}=\frac{5}{4}\)

⇒ \(\frac{32 x-29}{28}=\frac{5}{4}\)

⇒ 32x -29 = \(\frac{5}{4}\) x 28

⇒ 32x = 35 + 29

⇒ 32x = 64

⇒ x = \(\frac{64}{32}\)

⇒ x = 2

7. 3(x – 5)² + 5x = (2x – 3)² – (x + 1)² + 1

Solution.

Given

⇒ 3(x² – 10x + 25) + 5x = 4x² – 12x + 9 – x² – 2x – 1 + 1

⇒ 3x² – 30x + 75 + 5x = 3x² – 14x + 9

⇒ 3x² – 30x + 5x – 3x² + 14x = 9 – 75

⇒ -11x = -66

⇒ 11x = 66

⇒ x = \(\frac{66}{11}\)

⇒ x = 6

8. \(\frac{y+1}{2}-\frac{5 y+9}{28}=\frac{y+6}{21}+5-\frac{y-12}{3}\)

Solution.

Given

⇒ \(\frac{14 y+14-5 y-9}{28}=\frac{y+6+105-7 y+84}{21}\)

⇒ \(\frac{9 y+5}{28}=\frac{195-6 y}{21}\)

⇒ \(\frac{9 y+5}{4}=\frac{195-6 y}{3}\) [Multiplying both side by 7]

⇒ 3(9y + 5) = 4(195 – 6y)

⇒ 27y + 15 = 780 – 24y

⇒ 27y + 24y = 780 – 15

⇒ 51y = 765

⇒ y = \(\frac{765}{51}\) = 15

⇒ y = 15

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

Solution.

Given

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

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

⇒ \(\frac{1}{x(x+1)}=\frac{1}{(x+2)(x+3)}\)

⇒ \((x+2)(x+3)=x(x+1)\)

⇒ \(x^2+3 x+2 x+6=x^2+x\)

⇒ \(x^2+5 x-x^2-x=-6\)

⇒ 4x = -6

⇒ x = \(-\frac{6^3}{4^2}\)

⇒ x = \(-\frac{3}{2}\)

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

Solution.

Given

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

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

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

⇒ \(\frac{3 x-12-3 x+6}{(x-2)(x-4)}\)

⇒ \(\frac{4 x-12-4 x+16}{(x-4)(x-3)}\)

⇒ \(\frac{-6}{(x-2)(x-4)}=\frac{4}{(x-4)(x-3)}\)

⇒ \(\frac{-6}{x-2}=\frac{4}{x-3}\) [Multiplying both side by (x-4)]

⇒ 4(x – 2) = 6(x – 3) ⇒ 4x – 8 =- 6x + 18 ⇒ 4x + 6x = 18 + 8

⇒ 10x = 26

⇒ x = \(\frac{26}{10}\)

⇒ x = \(\frac{13}{5}\)

Formation of an equation:

Example 2. Find a number such that one-third of the number is 7 more than one-fourth of the number.

Solution: Let the number be x.

According to condition,

\(\frac{x}{3}-\frac{x}{4}=7 \Rightarrow \frac{4 x-3 x}{12}=7\)

⇒ \(\frac{x}{12}\) = 7

⇒ x = 12 x 7

⇒ x = 84

Hence the required number is 84.

Example 3. Divide 830 into two parts in such a way, that 30% of one part will be 4 more than 40% of the other.

Solution: Let 1st part be x

∴ 2nd part is (830 – x)

According to the question,

⇒ x x 30% = (830 – x) x 40% + 4

⇒ \(x \times \frac{30}{100}=(830-x) \times \frac{40}{100}+4\)

⇒ \(\frac{3 x}{10}=(830-x) \times \frac{4}{10}+4\)

⇒ \(\frac{3 x}{10}=\frac{3320-4 x+40}{10}\)

⇒ 3x = 3360 – 4x [Multiplying both side by 10]

⇒ 3x + 4x = 3360

⇒ 7x = 3360

⇒ \(x=\frac{3360}{7}\)

⇒ x = 480

∴ One part is 480 and the other part is (830 – 480) or 350.

Example 4. Find a fraction whose denominator is 2 more than the numerator and when 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes equal to \(\frac{7}{3}\)

Solution: Let the numerator of the fraction be x.

∴ Denominator is (x + 2)

∴ fraction is \(\frac{x}{x+2}\)

According to the conditions,

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

⇒ 7x – 3x = 9 + 7

⇒ 4x = 16

⇒ x = \(\frac{16}{4}\) = 4

∴ The required fraction is \(\frac{4}{4+2} \text { or } \frac{4}{6}\)

Example 5. A number consists of two digits whose sum is 9. If 27 is added to the number, the digits are interchanged. Find the number.

Solution: Let the unit’s digit be x.

Then ten’s digit = (9 – x)

∴ Number = 10(9 – x) + x = 90 – 10x + x = 90 – 9x

If the digits are interchanged then the number will be 10x + (9 − x) = 9x + 9

According to the condition,

9x + 9 = (90 – 9x)+ 27

= 9x + 9x = 90 + 27 – 9

⇒ 18x = 108

⇒ x = \(\frac{108}{18}\) = 6

∴ The required number is 90 – 9 x 6 = 90 – 54 = 36

Example 6. A father is thrice as old as his son. Four years back the father was 4 times as old his son. Find their present ages.

Solution: Let the present age of son is x years.

Father’s age is 3x years.

Four years ago son age was (x-4) years and father age was (3x – 4) years.

According to condition,

4(x – 4) = 3x – 4

⇒ 4x – 16 = 3x – 4

⇒ 4x – 3x = 16 – 4

⇒ x = 12

∴ The present age of son is 12 years and father’s age is (12 x 3) years or 36 years.

Example 7. Ramesh went to the school at a speed of 4 km/hr and returned at 3 km/hr. If he took 30 minutes more in returning, find the distance of the school from his home.

Solution: Let the distance between the school and home is x km.

Time taken in going x km from home to school is \(\frac{x}{4}\) hr.

Time taken in going x km from school to home is \(\frac{x}{3}\) hr.

According to condition,

\(\frac{x}{3}-\frac{x}{4}=\frac{30}{60}\left[30 \text { mins }=\frac{30}{60} \mathrm{hr}\right]\)

⇒ \(\frac{4 x-3 x}{12}=\frac{1}{2} \Rightarrow x=\frac{12}{2}=6\)

∴ The distance of the school from his home is 6 km.

Example 8. The length of a rectangle is 4 cm more than its breadth. If the perimeter of the rectangle is 36 cm, find its area.

Solution: Let the breadth of the rectangle is x cm.

The length is (x + 4) cm.

The perimeter is 2 (x + 4 + x) cm = 2 (2x + 4) cm.

According to question,

2(2x + 4) = 36

⇒ 4x + 8 = 36

⇒ 4x = 36 – 8 = 28

⇒ x = \(\frac{28}{4}\) = 7

∴ The breadth is 7 cm and length is (7 + 4) cm or 11 cm

Area = (11 x 7) cm² = 77 cm².

Example 9. Choose the correct answer:

1. If average of (x + 3) and (x – 7) is 5, then the value of x is

1. 5
2. 7
3. 6
4. 4

Solution: \(\frac{x+3+x-7}{2}=5\)

⇒ 2x – 4 = 10

⇒ 2x = 14

⇒ x = 7

∴ The correct answer is 2. 7

The value of x is 2. 7

2. What is the equation?

1. x + 2 = 3
2. (x + 2)² = x² + 4x + 4
3. x + 2
4. None of these

Solution: The correct answer is 1. x + 2 = 3

3. In the expression px + qy + r = 0, if x = 0 then the value of y is

1. \(\frac{r}{q}\)
2. \(\frac{r}{p}\)
3. 0
4. –\(\frac{r}{q}\)

Solution: px + qy + r = 0

∴ p x 0 + qy + r = 0

⇒ qy = -r ⇒ y = –\(\frac{r}{q}\)

∴ The correct answer is 4. –\(\frac{r}{q}\)

The value of y is –\(\frac{r}{q}\)

Example 10. Write ‘True’ or ‘False’:

1. If \(\frac{x}{9}\) = 0, then x = 9.

Solution: The statement is false.

2. The root of the equation 2x + 5 = 7x – 45 is 10.

Solution: 2x + 5 = 7x – 45

⇒ 2x – 7x = 45 – 5

⇒ -5x = – 50

⇒ 5x = 50

⇒ x = 10

∴ The statement is true.

3. If a(x + a) = b(x + b), then x = -(a + b)

Solution: a(x + a) = b(x + b)

⇒ ax + a² = bx + b²

⇒ ax – bx = b² – a²

⇒ x(a – b) = -(a² – b²)

⇒ x = \(-\frac{(a+b)(a-b)}{(a-b)}\)

⇒ x = −(a + b)

∴ The statement is true.

Example 11. Fill in the blanks:

1. __________ is the root of the equation 3x + 5 = 0.

Answer: 3x + 5 = 0

⇒ 3x = -5

⇒ x = –\(\frac{5}{3}\)

2. (x + a)² = x² + 2ax + a² is a __________

Answer: Identity.

3. If \(\frac{x}{2}\) + \(\frac{1}{3}\) = \(\frac{x}{3}\) + \(\frac{1}{2}\), then x = __________

Answer: \(\frac{x}{2}+\frac{1}{3}=\frac{x}{3}+\frac{1}{2}\)

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

⇒ \(\frac{3 x-2 x}{6}=\frac{3-2}{6} \Rightarrow \frac{x}{6}=\frac{1}{6}\)

⇒ x = 1.