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

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