## Algebra Chapter 12 Equations

**Equations Introduction**

In class VII you have learned about the formation of linear equations of one variable and their solutions. Formation of equations, some useful information about equations, and rules of solving an equation have been discussed in detail in the chapter of ‘equation^{1} in the book for class VII. In this chapter, our aim is to deal with a little harder problems and also to discuss some special techniques for solving algebraic equations involving one variable.

**Class 8 Maths Solutions Wbbse**

**Method of Transportation**

In an equation, usually, there are some terms on the left-hand side of the ‘=’ sign and also there are some terms on the right-hand side of the sign. If any term is taken from the left-hand side to the right-hand side or from the right-hand side to the left-hand side then a change of sign of the respective term takes place. This is called the method of transportation.

For example, if x + a = b then x = b – a, and if x + c = d then x + c – d = 0

**Cross multiplication**

If x/a = b/c then cx =ab. This process of multiplication of the numerator of one side of an equation by the denominator of the other side is called cross multiplication.

**Method Of Alternends**

If a/b = c/d then a/c = b/d.

This is called the method of alternates.

**Method Of Break Term**

In order to solve the equation of the form \(\frac{p}{x+a}+\frac{q}{x+b}=\frac{p+q}{x+c}\) it is convenient to proceed by writing the equation as

\(\frac{p}{x+a}+\frac{q}{x+b}=\frac{p}{x+c}+\frac{q}{x+c}\) Then is known as the method of break term.

For example, to solve the equation, \(\frac{2}{x+1}+\frac{3}{x+2}=\frac{5}{x+3}\)

write the equation as: 2 / x+ 1 + 3 / x + 2 = 2/ x + 3 + 3 / x + 3.

**Read And Learn More WBBSE Solutions For Class 8 Maths**

**Method of division**

In some cases, each fraction of an equation is written in such a way that, a part of the numerator becomes divisible by the denominator.

For example, to solve the equation,

\(\frac{4 x+5}{2 x-3}+\frac{3 x+8}{x-5}=\frac{15 x}{3 x-1}\)**Class 8 Maths Solutions Wbbse**

**we may proceed as follows:**

## Algebra Chapter 12 Equations Some examples of equations

Example 1

**Solve the equation : 5(x – 2) +7(x – 3) = 2(x + 3) + 3.**

**Solution :**

**Given 5(x – 2) +7(x – 3) = 2(x + 3) + 3.**

5(x – 2) + 7(x – 3) = 2(x + 3) + 3

or, 5x – 10 + 7x – 21 = 2x + 6 + 3

or, 12x – 31 = 2x + 9

or, 12x – 2x = 31 + 9

or, 10x = 40

or, x = 40/10

= 4

The required solution is x = 4.

**Class 8 Maths Solutions Wbbse**

Example 2

**Solve: (x – 2)(x – 3) = x ^{2} – 44.**

**Solution:**

(x – 2)(x – 3) = x^{2} – 44

or, x^{2} – 3x -2x + 6 = x^{2} – 44

or, x^{2} – x^{2} – 5x = -44 – 6

or, -5x = -50

or, x = -50 / -5

= 10

The required solution is x = 10.

Example 3

**Solve: (x -+ 1) + x(x – 3) = 2x ^{2} – 8.**

**Solution:**

**Given**

**(x -+ 1) + x(x – 3) = 2x ^{2} – 8.**

x(x + 1) + x(x – 3) = 2x^{2} – 8

or, x^{2} + x + x^{2} – 3x = 2x^{2} – 8

or, 2x^{2} – 2x^{2} – 2x = -8

or, -2x = -8

or, x = -8 / -2

= 4

The required solution is x = 4.

**Example 4**

**Solve: \(\frac{x+1}{5}+x=\frac{2 x+7}{5}+2\)**

**Solution:**

**Given**

** \(\frac{x+1}{5}+x=\frac{2 x+7}{5}+2\).**

\(\frac{x+1}{5}+x=\frac{2 x+7}{5}+2\)

\(\text { or, } \frac{x+1+5 x}{5}=\frac{2 x+7+10}{5}\)

\(\text { or, } 6 x+1=2 x+17 \text { or, } 6 x-2 x=17-1\)

\(\text { or, } 4 x=16\)

\(\text { or, } x=\frac{16}{4}=4\)

**Example 5**

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

**Solution:**

**Given \(\frac{2x+1}{7}+4=\frac{5-x}{3}\)**

**Ganit Prabha Class 8 Solution**

**Example 6**

**Solve: \(\frac{x+12}{6}-x=6 \frac{1}{2}-\frac{x}{12} \ldots\)**

**Solution:**

**Given \(\frac{x+12}{6}-x=6 \frac{1}{2}-\frac{x}{12} \ldots\).**

**Example 7**

**Solve: \(\frac{x-3 a}{b}+\frac{x-3 b}{a}+\frac{x-9 a-9 b}{a+b}=0 .\)**

**Solution:**

**Given \(\frac{x-3 a}{b}+\frac{x-3 b}{a}+\frac{x-9 a-9 b}{a+b}=0 .\)**

Now, the left-hand side is the product of two expressions. Its value will be equal to zero if at least one of them is zero. Clearly, the second factor cannot be equal to zero because nothing is known about the values of a, b, and c.

∴ x – 3a – 3b = 0

or, x = 3a +3b

The required solution is x = 3a +3b.

**Ganit Prabha Class 8 Solution**

**Example 8**

**Solve : \(\frac{3}{5}(x-4)-\frac{1}{3}(2 x-9)=\frac{1}{4}(x-1)-2\)**

**Solution:**

**Given**

Multiplying both sides by 60, which is the L.C.M of 5, 3, and 4.

36(x – 4) – 20(2x – 9) = 15(x – 1) – 120

or, 36x – 144 – 40x + 180 = 15x -15 -120

or, – 4x + 36 = 15x -135

or, 15x + 4x = 135 + 36

or, 19x = 171

or, x = 171 /19

= 9

The required solution is x = 9.

**Example 9**

**Solve: \(\frac{x+1}{2}-\frac{5 x+9}{28}=\frac{x+6}{21}+5-\frac{x-12}{3}\)**

**Solution:**

**Given \(\frac{x+1}{2}-\frac{5 x+9}{28}=\frac{x+6}{21}+5-\frac{x-12}{3}\)**

**Ganit Prabha Class 8 Solution**

**Example 10**

**Find the value of x from equation, \(\frac{7 x+5}{15}+\frac{2 x-5}{5}=\frac{2 x+4}{6}+\frac{5 x-2}{12}\)**

**Solution:**

**Given:-**

**\(\frac{7 x+5}{15}+\frac{2 x-5}{5}=\frac{2 x+4}{6}+\frac{5 x-2}{12}\)**

Example 11

**Solve: \(\frac{x^2+2 x+3}{x^2+3 x+5}=\frac{x+2}{x+3}\)**

**Solution:**

**Given:-**

**\(\frac{x^2+2 x+3}{x^2+3 x+5}=\frac{x+2}{x+3}\)**

**Example 12**

**Solve: \(\frac{2 x+1}{5}+\frac{7 x+2}{3}=\frac{x+3}{10}+\frac{2 x+1}{21}\)**

**Solution:**

**Given:-**

**\(\frac{2 x+1}{5}+\frac{7 x+2}{3}=\frac{x+3}{10}+\frac{2 x+1}{21}\)**

**Example 13**

**Solve: \(\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{1}{x-1} .\)**

**Solution:**

**Given:-**

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

**Example 14**

**Solve: \(\frac{3}{x-1}+\frac{2}{x-2}=\frac{5}{x-3}\)**

**Solution:**

**Given:-**

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

The required equation is 1 3/4

**Example 15**

**Solve: \(\frac{1}{x-2}+\frac{1}{x-6}=\frac{1}{x-3}+\frac{1}{x-5}\)**

**Solution:**

**Given :-**

**\(\frac{1}{x-2}+\frac{1}{x-6}=\frac{1}{x-3}+\frac{1}{x-5}\)**

**Ganit Prabha Class 8 Solution**

**Example 16**

**Find the value of x from the relation \(\frac{a}{a-x}+\frac{b}{b-x}=\frac{a+b}{a+b-x} .\)**

**Solution:**

**Given:-**

**\(\frac{a}{a-x}+\frac{b}{b-x}=\frac{a+b}{a+b-x} .\)**

From the given relation,

**Example 17**

**Solve: \(\begin{aligned}
\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)} & +\frac{1}{(x+3)(x+4)} \\
& =\frac{1}{(x+1)(x+6)}
\end{aligned}\)**

**Solution:**

From the given equation,

**Ganit Prabha Class 8 Solution**

**Example 18**

**Solve: \(\frac{2 x+1}{2 x-1}+\frac{3 x+1}{3 x-2}=\frac{3 x+2}{3 x-1}+\frac{6 x+1}{6 x-5}\)**

**Solution:**

From the given equation,

**Example 19**

**Solve: \(\frac{x-b c}{b+c}+\frac{x-c a}{c+a}+\frac{x-a b}{a+b}=a+b+c\)**

**Solution:**

From the given equation,

**Maths Solutions Class 8 Wbbse**

Now, the left-hand side is the product of two quantities. Its value will be equal to zero if at least one of them is zero. Clearly, the second factor cannot be equal to zero because nothing is known about the values of a, b, and c.

Hence, x – ab – be – ca = 0

or, x = ab + bc + ca

The required solution is x = ab + bc + ca.

Example 20

**Solve: \(\frac{x+a^2+2 c^2}{b+c}+\frac{x+b^2+2 a^2}{c+a}+\frac{x+c^2+2 b^2}{a+b}=0 .\)**

**Solution:**

From the given equation,

\(\begin{array}{r}\frac{x+a^2+2 c^2}{b+c}+(b-c)+\frac{x+b^2+2 a^2}{c+a}+(c-a) \\

+\frac{x+c^2+2 b^2}{a+b}+(a-b)=0

\end{array}\)

Now, the left-hand side is the product of two quantities. Its value will be equal to zero if at least one of them is zero’. Clearly, the second factor cannot be equal to zero because nothing is known about the values of a, b, and c.

Hence, x + a^{2} + b^{2} + c^{2} = 0

or, x = – (a^{2} + b^{2} + c^{2})

The required solution is x = – (a^{2}+ b^{2} + c^{2})

Example 21

**The sum of the digits of a number of two digits is 10. if the digit in the units’ place is 4 times the digit in the tens’ place, find the number.**

**Solution:**

Let, the digit in the tens place be x.

Then the digit in the units place = 4x

∴ The number = 10 x x + 4x

= 10x + 4x

= 14x.

Again, according to the question,

x + 4x = 10

or, 5x = 10

or, x = 10/5

= 2

Hence, the number = 14 x 2

= 28

The number is 28.

**Maths Solutions Class 8 Wbbse**

Example 22

**A number consisting of two digits is such that its digit in the tens’ place is twice that in the units’ place. If the digits are reversed, the number thus formed is 27 less than the original one. Find the number.**

**Solution :**

Let, the digit in the units’ place be x.

Then the digit in the tens’ place = 2x.

∴ The number = 10 x 2x + x = 20x + x= 21x

If the digits are reversed, the number obtained

= 10 x x + 2x

= 12x

According to the question,

21x – 12x = 27

or, 9x = 27

or, x = 27/9

= 3

Hence, the number

= 21 x 3

= 63

The number is 63.

Example 23

**The denominator of a fraction is greater than its numerator by 2. The fraction becomes 1/2 when 3 is subtracted from both the numerator and the denominator. Find the fraction.**

**Solution:**

Let, the numerator of the fraction be x.

Then its denominator = x + 2.

**Maths Solutions Class 8 Wbbse**

Example 24

**The denominator of a fraction is 2 more than twice the numerator keeping the denominator unaltered then the fraction becomes 7/12. Find the fraction.**

**Solution:**

Let, the numerator be x.

**Maths Solutions Class 8 Wbbse**

Example 25

* ₹* 624 was distributed among A, B, C, and D, If A would receive ₹ 2 more, B would receive

*6 less, C would receive 5 times his money and D would receive 1/4th of his money, then they would receive equal money. How much money did each of them receive?*

*₹***Solution:**

**Given**

* ₹* 624 was distributed among A, B, C, and D, If A would receive ₹ 2 more, B would receive

*6 less, C would receive 5 times his money and D would receive 1/4th of his money, then they would receive equal money.*

*₹*Let, the equal money be ₹ x,

Then A received ₹( x – 2),

B received ₹ ( x + 6),

C received ₹ x/5 ,

D received ₹ 4x.

According to the question,

x – 2 + x + 6 + x/5 + 4x = 624

or, 6x + x/5 = 624 – 4

or, 31x / 5 = 620

or, x = 620 x 5/31 = 100

∴ A received ₹(100-2)

= ₹98

B received ₹ (100+6)

= ₹ 106

C received ₹ 100/5

= ₹ 20

D received ₹ 4 x 100

= ₹ 400.

∴ A received v 98,

B received ₹ 106,

C received ₹ 20,

D received ₹ 400.