Linear Equations in One Variable: Exercise 6

We use cross multiplication in this exercise a lot of times, so it is explained here before.

Let, a/b = c/d Now if we multiply both sides by the denominators of left side and right side, we get, (a/b) X (b X d) = (c/d) X (b X d) => a X d = b X c This is called cross multiplication.

Question.1 Solve the following equations.

We can solve the problems 1 to 5 by trying to bring all the unknown variables to the left side.

1. (8x-3) / 3x = 2

Solution:

By multiplying on both sides by 3x we get, => (8x-3) X (3x) / 3x = 2 X (3x) => 8x-3 = 6x => 8x-6x-3 = 0 => 2x-3 = 0 => 2x = 3 => x = 3/2

2. 9x / (7-6x) = 15

Solution:

By multiplying both sides by (7-6x) we get, => (9x) X (7-6x) / (7-6x) = 15 X (7-6x) => 9x = (15 X 7) – (15 X 6)x => 9x = 105 – 90x => 9x + 90x = 105 => 99x = 105 => x = 105/99 => x = 35/33

3. z / (z+15) = 4 / 9

Solution:

By cross multiplication, => z X 9 = (z+15) X 4 => 9z = 4z + 4 X 15 => 9z – 4z = 60 => 5z = 60 => z = 60/5 => z = 12

4. (3y+4) / (2-6y) = 2 / 5

Solution:

By cross multiplication, => (3y+4) X 5 = (-2) X (2-6y) => (5 X 3)y + (4 X 5) = (-2 X 2) + (-2 X -6)y => 15y + 20 = -4 + 12y => 15y -12y = (-4) + (-20) => 3y = -24 => y = -24/3 => y = -8

5. (7y+4) / (y+2) = – 4 / 3

Solution:

By cross multiplication, => (7y+4) X 3 = -4 X (y+2) => (7 X 3)y + (4 X 3) = -4y + (-4 X 2) => 21y + 12 = (-4y) + (-8) => 21y + 4y = (-8) + (-12) => 25y = -20 => y = -20/25 => y=-4/5

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 age of Hari be x , Let the present age of Harry be y . Presently their ages are in ratio 5:7 , So we get => x : y = 5 : 7 We get, => x/y = 5/7 By cross multiplication, => 7x = 5y => x = (5y) / 7 ……….. (1) After 4 years, Hari’s age will be x+4, Harry’s age will be y+4. The ratio between their ages after four years is 3:4. So we get, => (x+4) : (y+4) = 3 : 4 => (x+4)/(y+4) = 3/4 By cross multiplication, => (x+4) X 4 = 3 X (y+4) => 4x +16 = 3y + 12 => 4x – 3y = -4 ……….. (2) Now , we got two euations. x = (5y) / 7 ……….. (1) 4x – 3y = -4 ……….. (2) If we substitute this x value from (1) in equation (2) we get => 4 X (5y/7) – 3y = -4 => 20y/7 – 3y = -4 => 20y/7 – (7X3)y/7 = -4 => (20y -21y) / 7 = -4 => -y/7 = -4 => y = (-4) X (-7) => y = 28 By substituting y=28 value in (1) we get => x = (5 X 28) / 7 => x = (5 X 4) => x =20 So here Hari’s present age is 20 years and Harry’s present age is 28 years.

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 3/2. Find the rational number.

Solution:

Let the numerator be x, and the denominator be y. From the first part of the question we get, => denominator = numerator + 8 => y = x + 8 ……….. (1) Now, from the second part of the question , => (x+17) / (y-1) = 3/2 By cross multiplication, => (x+17) X 2 = 3 X (y-1) => 2x + 34 = 3y – 3 => 2x – 3y = -34 – 3 => 2x – 3y = -37 ……….. (2) We got two equations. Substituting (1) in (2), we get => 2x − 3 X (x+8) = −37 => 2x − 3x − 24 = −37 => 37 − 24 = x => x = 13 By substituting x=13 in (1) we get, => y = 13 +8 => y = 21 We got x=13 and y=21 Hence the original rational fraction will be 13/21

NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable is created by a team of professionals at GFG, to help students with any queries they might have as they go through problems from the NCERT textbook. It lessens the frustration of spending a long time working on a problem.

Chapter 2 of Class 8 Maths covers the topic of linear equations in one variable. Including topics such as:

• Linear Equations
• Solving Equations which have Linear Expressions on one Side and Numbers on the other Side
• Solve Linear Equations with Variable on both Sides
• Reducing Equations to Simpler Form
• Equations Reducible to Linear Form