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

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

PSEB 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:
\(\frac{8 x-3}{3 x}=2\)
∴ 3x\(\left(\frac{8 x-3}{3 x}\right)\) = 3x (2) (Multiplying both the sides by 3x)
∴ 8x – 3 = 6x
∴ 8x – 6x = 3 [Transposing 6x to LHS and (-3) to RHS]
∴ 2x = 3
∴ \(\frac{2 x}{2}=\frac{3}{2}\) (Dividing both the sides by 2)
∴ 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) (Cross multiplication)
∴ 9x = 105 – 90x
∴ 9x + 90x = 105 [Transposing (- 90x) to LHS]
∴ 99x = 105
∴ \(\frac{99 x}{99}=\frac{105}{99}\)(Dividing both the sides by 99)
∴ x = \(\frac {35}{33}\)

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

Question 3.
\(\frac{z}{z+15}=\frac{4}{9}\)
Solution:
\(\frac{z}{z+15}=\frac{4}{9}\)
∴ z(9) = 4(z + 15) (Cross multiplication)
∴ 9z = 4z + 60
∴ 9z – 4z = 60 (Transposing 4z to LHS)
∴ 5z = 60
∴ \(\frac{5 z}{5}=\frac{60}{5}\) (Dividing both the sides by 5)
∴ 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) (Cross multiplication)
∴ 15y + 20 = -4 + 12y
∴ 15y – 12y = – 4 – 20 (Transposing 12y to LHS and 20 to RHS)
∴ 3y = -24
∴ \(\frac{3 y}{3}=\frac{-24}{3}\) (Dividing both the sides by 3)
∴ 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) (Cross multiplication)
∴ 21y + 12 = – 4y – 8
∴ 21y + 4y = – 8 – 12 (Transposing -4y to LHS and 12 to RHS)
∴ 25y = -20
∴ \(\frac{25 y}{25}=\frac{-20}{25}\) (Dividing both the sides by 25)
∴ y = \(\frac {-4}{5}\)

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

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:
Age of Hari : Age of Harry
= 5 : 7
Let the present age of Hari be 5x years.
Then, the present age of Harry = 7x years.
After 4 years their ages :
Hari = (5x + 4) years
Harry = (7x + 4) years
∴ (5x + 4) : (7x + 4) = 3 : 4
∴ \(\frac{5 x+4}{7 x+4}=\frac{3}{4}\)
∴ 4(5x + 4) = 3(7x + 4) (Cross multiplication)
∴ 20x + 16 = 21x + 12
∴ 20x – 21x = 12 – 16 (Transposing 21x to LHS and 16 to RHS)
∴ -x = – 4
∴ x = 4 [Multiplying both the sides by (- 1)]
∴ Hari’s present age = 5x = 5 × 4
= 20 years
∴ Harry’s present age = 7x = 7 × 4
= 28 years
Thus, Hari’s present age is 20 years and Harry’s present age is 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 be x.
Denominator (8 more than numerator) = x + 8
New numerator = x + 17
(After adding 17)
New denominator = x + 8 – 1
= x + 7
(After decreasing 1)
But new number = \(\frac{x+17}{x+7}\)
But this rational number is \(\frac {3}{2}\)
\(\frac{x+17}{x+7}=\frac{3}{2}\)
∴ 2(x + 17) = 3(x + 7) (Cross multiplication)
∴ 2x + 34 = 3x + 21
∴ 2x – 3x = 21 – 34 (Transposing 3x to LHS and 34 to RHS)
∴ -x = – 13
∴ x = 13 [Multiplying both the sides by (-1)]
∴ Numerator = x = 13
Denominator = x + 8
= 13 + 8
= 21
The rational number = \(\frac {13}{21}\)
Thus, the rational number is \(\frac {13}{21}\).

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