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

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 Textbook Exercise Questions and Answers.

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

Question 1.
Solve the following equations
\(\frac{8 x-3}{3 x}\) = 2
Answer:
Given, \(\frac{8 x-3}{3 x}\) = 2
Multiplying both sides by 3x,
3 × \(\left(\frac{8 x-3}{3 x}\right)\) = 2 × 3x
or 8x - 3 = 6x
or 8x - 6x = 3
or 2x = 3
or x = \(\frac{3}{2}\)

Question 2.
\(\frac{9 x}{7-6 x}\) = 15
Answer:
Given, \(\frac{9 x}{7-6 x}\) = 15
Multiplying both sides by 7 - 6x,
(7 - 6x) × \(\frac{9 x}{7-6 x}\) =(7 - 6x) × 15
or 9x = 105 - 90x
or 9x + 90x = 105
or 99x = 105
or x = \(\frac{105}{99}=\frac{35}{33}\)

Question 3.
\(\frac{z}{z+15}=\frac{4}{9}\)
Answer:
Given \(\frac{z}{z+15}=\frac{4}{9}\)
MuItip)yin both sides by 9(z + 15).
9(z+15) × \(\frac{z}{z+15}\) = 9(z + 15) × \(\frac{4}{9}\)
or 9z =4z + 60
or 9z - 4z = 60
or 5z = 60
or z = \(\frac{60}{5}\) = 12

Question 4.
\(\frac{z}{z+15}=\frac{4}{9}\)
Answer:
Given \(\frac{z}{z+15}=\frac{4}{9}\)
Cross-rnultiplication gives,
or 5(3y + 4) = -2(2 - 6y)
or 15y + 20 = -4 + 12y
or 15y - 12y = -4 - 20
or 3y = - 24
or y = \(\frac{-24}{3}\)
or y = - 8

Question 5.
\(\frac{7 y+4}{y+2}=\frac{-4}{3}\)
Answer:
Given \(\frac{7 y+4}{y+2}=\frac{-4}{3}\)
Cross-multiplication gives,
or 3(7y + 4) = - 4(y + 2)
or 21y + 12= - 4y - 8
or 21y + 4y= - 8 - 12
or 25y= - 20
or y = \(\frac{-20}{25}\)
or 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. 
Answer:
Let the present age of Hari and Harry be 5.x years and 7x years respectively
After 4 years - Hari’s age = (5x + 4) years
Harry’s age = (7x + 4) years
Given that \(\frac{5 x+4}{7 x+4}\) = \(\frac{3}{4}\)

By cross multiplication -
4(5x + 4) = 3(7x + 4)
or 20x + 16= 21x + 12
or 20x - 21x= 12 - 16
or - x = - 4, i.e. x = 4
Hari’s present age = 5x = 5 × 4 years
= 20 years
And Harry’s present age = 7x = 7 × 4 years
= 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.
Answer:
Let the numerator of the rational number be x.
Then, the denominator of the rational number = x + 8
The rational number = \(\frac{x}{x+8}\)
If the numerator is increased by 17 and the denominator is decreased by 1, the number becomes \(\frac{3}{2}\)
∴ \(\frac{x+17}{(x+8)-1}=\frac{3}{2}\)
or \(\frac{x+17}{x+7}=\frac{3}{2}\)
or 2(x + 17) = 3(x + 7)
or 2x + 34 = 3x + 21
or 2x - 3x= 21 - 34
or x= 13
∴ \(\frac{x}{x+8}=\frac{13}{13+8}=\frac{13}{21}\)
Hence the required rational number = \(\frac{13}{21}\)