You can Download KSEEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6

Solve the following equations:

Question 1.

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

Solution:

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

⇒ 8x – 3 = 6x

⇒ 8x – 6x = 3

⇒ 2x = 3

⇒ x = \(\frac{3}{2}\)

Question 2.

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

Solution:

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

⇒ 9x = 15(7 – 6x)

⇒ 9x = 105 – 90x

⇒ 9x + 90x = 105

⇒ 99x = 105

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

⇒ x = \(\frac{35}{33}\)

Question 3.

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

Solution:

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

⇒ 9z = 4(z + 15)

⇒ 9z = 4z + 60

⇒ 9z – 4z = 60

⇒ 5z = 60

⇒ z = 12

Question 4.

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

Solution:

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

⇒ 5(3y + 4) = -2(2 – 6y)

⇒ 15y + 20 = -4 + 12y

⇒ 15y – 12y = -4 – 20

⇒ 3y = -24

⇒ y = -8

Question 5.

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

Solution:

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

⇒ 3(7y + 4) = -4(y + 2)

⇒ 21y + 12 = -4y – 8

⇒ 21y + 4y = -8 – 12

⇒ 25y = -20

⇒ 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 = 5x years

Harry’s present age = 7x years

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

Harry’s age = (7x + 4) years

According to question,

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

⇒ 4(5x + 4) = 3(7x + 4)

⇒ 20x + 16 = 21x + 12

⇒ 16 – 12 = 21x – 20x

⇒ 4 = x

∴ 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 number obtained is \(\frac{3}{2}\). Find the rational number.

Solution:

Let the numerator of a fraction = x

Denominator of a fraction = x + 8

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

If numerator is increased by 17

then new numerator = (x + 17)

New denominator = x + 8 – 1 = (x + 7)

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

According to question,

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

⇒ 2(x + 17) = 3(x + 7)

⇒ 2x + 34 = 3x + 21

⇒ 34 – 21 = 3x – 2x

⇒ 13 = x

⇒ x = 13

Numerator = 13

Denominator = 13 + 8 = 21

Fraction = \(\frac{13}{21}\)