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

- Class 8 Maths Linear Equations in One Variable Exercise 2.1
- Class 8 Maths Linear Equations in One Variable Exercise 2.2
- Class 8 Maths Linear Equations in One Variable Exercise 2.3
- Class 8 Maths Linear Equations in One Variable Exercise 2.4
- Class 8 Maths Linear Equations in One Variable Exercise 2.5
- Class 8 Maths Linear Equations in One Variable Exercise 2.6
- Linear Equations in One Variable Class 8 Extra Questions

**NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.6**

**Solve the following equations.**

Ex 2.6 Class 8 Maths Question 1.

\(\frac { 8x-3 }{ 3x } =2\)

Solution:

We have \(\frac { 8x-3 }{ 3x } =2\)

⇒ \(\frac { 8x-3 }{ 3x }\) = \(\frac { 2 }{ 1 }\)

⇒ 8x – 3 = 2 × 3x (Cross-multiplication)

⇒ 8x – 3 = 6x

⇒ 8x – 6x = 3 (Transposing 6x to LHS and 3 to RHS)

⇒ 2x = 3

⇒ x = \(\frac { 3 }{ 2 }\)

Ex 2.6 Class 8 Maths Question 2.

\(\frac { 9x }{ 7-6x }\) = 15

Solution:

we have \(\frac { 9x }{ 7-6x }\) = 15

⇒ \(\frac { 9x }{ 7-6x }\) = \(\frac { 15 }{ 1 }\)

⇒ 9x = 15(7 – 6x) (Cross-multiplication)

⇒ 9x = 105 – 90x (Solving the bracket)

⇒ 9x + 90x = 105 (Transposing 90x to LHS)

⇒ 99x = 105

⇒ x = \(\frac { 105 }{ 99 }\)

⇒ x = \(\frac { 35 }{ 33 }\)

Ex 2.6 Class 8 Maths Question 3.

\(\frac { z }{ z+15 } =\frac { 4 }{ 9 }\)

Solution:

We have \(\frac { z }{ z+15 } =\frac { 4 }{ 9 }\)

⇒ 9z = 4 (z + 15) (Cross-multiplication)

⇒ 9z = 4z + 60 (Solving the bracket)

⇒ 9z – 42 = 60

⇒ 5z = 60

⇒ z = 12

Ex 2.6 Class 8 Maths Question 4.

\(\frac { 3y+4 }{ 2-6y } =\frac { -2 }{ 5 }\)

Solution:

we have \(\frac { 3y+4 }{ 2-6y } =\frac { -2 }{ 5 }\)

⇒ 5(3y + 4) = -2(2 – 6y) (Cross-multiplication)

⇒ 15y + 20 = -4 + 12y (Solving the bracket)

⇒ 15y – 12y = -4 – 20 (Transposing 12y to LHS and 20 to RHS)

⇒ 3y = -24 (Transposing 3 to RHS) -24

⇒ y = -8

Ex 2.6 Class 8 Maths Question 5.

\(\frac { 7y+4 }{ y+2 } =\frac { -4 }{ 3 }\)

Solution:

we have \(\frac { 7y+4 }{ y+2 } =\frac { -4 }{ 3 }\)

⇒ 3(7y + 4) = -4 (y + 2) (Corss-multiplication)

⇒ 21y + 12 = -4y – 8 [Solving the bracket]

⇒ 21y + 4y = -12 – 8 [Transposing 4y to LHS and 12 to RHS]

⇒ 25y = -20 [Transposing 25 to RHS]

⇒ y = \(\frac { -4 }{ 5 }\)

Ex 2.6 Class 8 Maths 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 the present ages of Hari and Harry be 5x years and 7x years respectively.

After 4 years Hari’s age will be (5x + 4) years and Harry’s age will be (7x + 4) years.

As per the conditions, we have

\(\frac { 5x+4 }{ 7x+4 } =\frac { 3 }{ 4 }\)

⇒ 4(5x + 4) = 3(7x + 4) (Cross-multiplication)

⇒ 20x + 16 = 21x + 12 (Solving the bracket)

⇒ 20x – 21x = 12 – 16 (Transposing 21x to LHS and 16 to RHS)

⇒ -x = -4

⇒ x = 4

Hence the present ages of Hari and Harry are 5 × 4 = 20years and 7 × 4 = 28years respectively.

Ex 2.6 Class 8 Maths 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 the rational number be x.

Denominator = (x + 8)

As per the conditions, we have

⇒ 2(x + 17) = 3(x + 7) (Cross-multiplication)

⇒ 2x + 34 = 3x + 21 (Solving the bracket)

⇒ 2x – 3x = 21 – 34 (Transposing 3x to LHS and 34 to RHS)

⇒ -x = -13

⇒ x = 13

Thus, numerator = 13

and denominator = 13 + 8 = 21

Hence the rational number is \(\frac { 13 }{ 21 }\).

#### More CBSE Class 8 Study Material

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- NCERT Solutions for Class 8 English It So Happened
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