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

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 = \(\frac{35}{33}\)

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

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

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

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.

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

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

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