NCERT MATHS SOLUTION CLASS 8 EXERCISE 2.6 | LINEAR EQUATIONS IN ONE VARIABLE

Find here step-by-step NCERT Maths Solution Class 8 Exercise 2.6 | Equations Reducible to the Linear Form | Chapter 2 Linear Equations In One Variable from CBSE NCERT textbook.

Before moving to maths solution class 8 exercise 2.6 , it is advised to go through the following topics:

• Linear Equations In One Variable

NCERT Maths Solution Class 8 Exercise 2.6

Problems in Exercise 2.6 require the knowledge of reducing non-linear equations (reducible) to linear equations.

Some equations may not be linear to begin with, but they can be brought to linear form by multiplying both sides of the equation by a suitable expression.[ NCERT MATHS SOLUTION CLASS 8 EXERCISE 2.6 ]

Cross Multiplication :

If an equation is in the form \frac{a}{b}=\frac{c}{d} we multiply both LHS and RHS with b and d but omit the step and write the result in next step as a\times d = b \times c}.

Exercise 2.6
Equations Reducible to the Linear Form

Solve the following equations.

1). \frac{8x-3}{3x}=2

2). \frac{9x}{7-6x}=15

3). \frac{z}{z+15}=\frac{4}{9}

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

5). \frac{7y+4}{y+2}=\frac{-4}{3}

SOLUTION

1). \frac{8x-3}{3x}=2

\implies 8x-3 = 2\times 3x

\implies 8x-3 = 6x

\implies 8x-6x = 3

\implies 2x = 3

\implies x = \frac{2}{3}

2). \frac{9x}{7-6x}=15

\implies 9x = 15\times (7-6x)

\implies 9x = 105-90x

\implies 9x+90x = 105

\implies 99x = 105

\implies x = \frac{105}{99}

\implies x = \frac{35}{33}

\implies x = 1\frac{2}{33}

3). \frac{z}{z+15}=\frac{4}{9}

\implies z\times 9 = 4\times (z+15)

\implies 9z = 4z+60

\implies 9z-4z = 60

\implies 5z = 60

\implies z = \frac{60}{5}

\implies z = 12

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

\implies (3y+4)\times 5 = (-2)\times (2-6y)

\implies 15y+20 = -4+12y

\implies 15y-12y = -4-20

\implies 3y = -24

\implies y = \frac{-24}{3}

\implies y = -8

5). \frac{7y+4}{y+2}=\frac{-4}{3}

\implies (7y+4)\times 3 = (-4)\times (y+2)

\implies 21y+12 = -4y -8

\implies 21y+4y = -8-12

\implies 25y = -20

\implies y = \frac{-20}{25}

\implies y = \frac{-4}{5}

6). The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of
their ages will be 3:4. Find their present ages.

SOLUTION

Let, the age of Hari be 5x years,
So, the age of Harry be 7x years,

Four years from now,

The age of Hari will be (5x + 4) years,
and the age of Harry will be (7x + 4) years,

A/Q,

\frac{5x+4}{7x+4}=\frac{3}{4}

\implies (5x+4)\times 4 = 3\times (7x+4)

\implies 20x+16 = 21x +12

\implies 20x - 21x = 12 - 16

\implies -x = -4

\implies x = 4

7). The denominator of a rational number is greater than its numerator by 8. If the
numerator is increased by 17 and the denominator is decreased by 1, the number
obtained is \frac{3}{2}. Find the rational number.

SOLUTION

Let, the numerator of the rational number be x,
So, the denominator of the rational number be x + 8,

A/Q,

\frac{x+17}{(x+8)-1}=\frac{3}{2}

\frac{x+17}{x+7}=\frac{3}{2}

\implies (x+17)\times 2 = 3\times (x+7)

\implies 2x+34 = 3x+21

\implies 2x - 3x = 21 - 34

\implies -x = -13

\implies x = 13

Therefore, the numerator of the rational number = x = 13,
and the denominator of the rational number = x + 8 = 21,

Hence, the rational number = \frac{13}{21}


We hope NCERT Maths Solution Class 8 Exercise 2.6 has helped you understand how to reduce non-linear equations to linear form and solve them.

Please write in the comment section for any error or any solution related queries from the exercise.

Check NCERT Solution of other Exercises from Class 8 Maths Chapter 2 Linear Equations in One Variable by clicking the links given below.


NCERT MATHS SOLUTION | CLASS 8 | CHAPTER 2 | LINEAR EQUATIONS IN ONE VARIABLE

Exercise 2.1 | Exercise 2.2 | Exercise 2.3

Exercise 2.4 | Exercise 2.5 | Exercise 2.6

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