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

   

Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6

Question 1.
Solve the following equations.
(i) \(\frac { 8x – 3 }{ 3x }\) = 2
(ii) \(\frac { 9x }{ 7 – 6x }\) = 15
(iii) \(\frac { z }{ z + 15 }\) = \(\frac { 4 }{ 9 }\)
(iv) \(\frac { 3y + 4 }{ 2 – 6y }\) = \(\frac { -2 }{ 5 }\)
(v) \(\frac { 7y + 4 }{ y + 2 }\) = \(\frac { -4 }{ 3 }\)
Solution:
(i) \(\frac { 8x – 3 }{ 3x }\) = 2
Multiplying both sides by 3x, we have
\(\frac { 8x – 3 }{ 3x }\) = 2 × 3x = 2 × 3x
or 8x – 3 = 6x
Transposing (-3) to RHS and 6x to LHS, we have
8x – 6x = 3 or 2x = 3
Dividing both sides by 2, we have x = \(\frac { 3 }{ 2 }\)

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

(ii) \(\frac { 9x }{ 7 – 6x }\) = 15
Multiplying both sides by 7 – 6x, we have
\(\frac { 9x }{ 7 – 6x }\) × (7 – 6x) = 15 × (7 – 6x)
or 9x = 105 – 90x
Transposing (-90x) to LHS, we have
9x + 90x = 105 or 99x = 105
or x = \(\frac { 105 }{ 99 }\) (Dividing both sides by 99)
or x = \(\frac { 35 }{ 33 }\)

(iii) \(\frac { z }{ z + 15 }\) = \(\frac { 4 }{ 9 }\)
by cross-multiplication, we have
9z = 4(z +15) ⇒ 9z = 4z + 60
Transposing 4z to LHS, we have
5z = 60 ⇒ z = \(\frac { 60 }{ 5 }\) = 12
∴ z = 12

(iv) \(\frac { 3y + 4 }{ 2 – 6y }\) = \(\frac { -2 }{ 5 }\)
By cross-multiplication, we have
5(3y + 4) = -2(2 – 6y) .
or 15y + 20 = – 4 + 12y
Transposing 20 to RHS and 12y to LHS, we
have
15y – 12y = – 4 – 20 or 3y = -24
or y = \(\frac { -24 }{ 3 }\) = -8
(Dividing both sides by 3) or y = -8

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

(v) \(\frac { 7y + 4 }{ y + 2 }\) = \(\frac { -4 }{ 3 }\)
By cross-multiplication, we have
3 x (7y + 4) = -4 x (y + 2)
or 21y + 12 = -4y – 8
Transposing 12 to RHS and (-4y) to LHS,
we have
21y + 4y = -8 – 12
or 25 y = -20 or y = \(\frac { -20 }{ 25 }\) = \(\frac { -4 }{ 5 }\)
(Dividing both sides by 25)
∴ y = \(\frac { -4 }{ 5 }\)

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

Question 2.
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 the present age of Hari = 5x years and the present age of Harry = 7x years After 4 years, age of Hari = (5x + 4) yearn Age of Harry = (7x + 4)’years According to the condition,
(5x + 4) : (7x + 4) = 3 : 4 or \(\frac { 5x + 4 }{ 7x + 4 }\) = \(\frac { 3 }{ 4 }\)
by cross multiplication, we have:
4(5x + 4) = 3(7x + 4)
or 20x + 16 = 21x + 12
Transposing 1+6 to RHS and 21x to LHS, we have
20x – 21x = 12 – 16 -x = -4 ⇒ x = 4
∴ Present age of Hari = 5 x 4 = 20 years
Present age of Harry = 7 x 4 = 28 years

Question 3.
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 = x
∴Denominator = x + 8
New numerator = x + 17
New denominator = (x + 8) – 1 = x + 1
∴ The new number = \(\frac { x + 17 }{ x + 7 }\)
According to the condition, we have
\(\frac { x + 17 }{ x + 7 }\) = \(\frac { 3 }{ 2 }\)
By cross multiplication, we have
2(x + 17) = 3(x + 7)
2x + 34 = 3x + 21
Transposing 34 to RHS and 3x to LHS, we have
2x – 3x = 21 – 34 ⇒ -x = -13 x = 13
∴ x = 13 ⇒ Numerator =13
x + 8 = 13 + 8 = 21⇒Denominator = 21
∴ The rational number = \(\frac { 13 }{ 21 }\)

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

Leave a Comment

Your email address will not be published. Required fields are marked *