Haryana State Board HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.6 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.6

Solve the following equations

Question 1.

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

Solution:

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

or, \(\frac{8 x-3}{3 x}\) 3x = 2 3x

or, 8x – 3 = 6x

or, 8x – 6x = 3

or, 2x = 3

or, x = \(\frac{3}{2}\)

∴ x = \(\frac{3}{2}\)

Question 2.

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

Solution:

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

or, \(\frac{9 x}{7-6 x}\) = 15

or, 15(7 – 6x) = 9x

or, 105 – 90x = 9x

or, 105 = 9x – 90x

or, 99x = 105

or, x = \(\frac{105}{99}\) = \(\frac{35}{33}\)

∴ x = \(\frac{35}{33}\)

Question 3.

\(\frac{z}{z+15}\) = \(\frac{4}{9}\)

Solution:

\(\frac{z}{z+15}\) = \(\frac{4}{9}\)

or, 9z = 4(z + 15)

or, 9z = 4z + 60

or, 9z – 4z = 60

or, 5z = 60

or, z = 12

∴ z = 12

Question 4.

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

Solution:

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

or, 5(3y + 4) = -2(2 – 6y)

or, 15y + 20 = -4 + 12y

or, 15y – 12y = -4 – 20

or, 3y = -24

or, y = \(\frac{-24}{3}\) = -8

∴ y = -8

Question 5.

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

Solution:

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

or, 3(7y + 14) = -4(y + 2)

or, 21y + 12 = -4y – 8

or, 21y + 4y = -8 – 12

or, 25y = -20

or, y = \(-\frac{20}{25}\) = \(-\frac{4}{5}\)

∴ y = \(-\frac{4}{5}\)

Question 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 Hari’s present age be 5x

Harry present age be 7x

After four years Hari’s age = (5x + 4) years

After four years Harry’s age = (7x + 4) years

According to the question

Therefore \(\frac{5x+4}{7x+4}\) = \(\frac{3}{4}\)

or, 4 (5x + A) = 3 (7x + 4)

or, 20x + 16 = 21x + 12

or, 20x – 21x = 12 – 16

or, -x = -4

or, x = 4

∴ Hari’s present age = 5 × 4 = 20 years.

Harry’s present age = 7 × 4 = 28 years.

Question 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 numberobtained is \(\frac{3}{2}\). Find the rational number.

Solution:

Let the numerator of a fraction be x

Denominator = x + 8

Fraction = \(\frac{x}{x+8}\)

2nd part of the question

Numerator = x + 17

Denominator = x + 8 – 1 = x + 7

Fraction = \(\frac{x+17}{x+7}\)

or, \(\frac{3}{2}\) = \(\frac{x+17}{x+7}\)

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

or, 3x + 21 = 2x + 34

or, 3x – 2x = 34 – 21

or, x = 13

(∴ denominator = 13 + 8 = 21)

∴ Fraction = \(\frac{13}{21}\)