Linear Equations l Exercise 3.3 l Class 10 l NCERT Solutions

1. Solve the following pair of linear equations

(i) x + y = 5 and 2x − 3y = 4

Step 1: From x + y = 5

x = 5 − y

Step 2: Substitute in second equation:

2(5 − y) − 3y = 4

10 − 2y − 3y = 4

10 − 5y = 4

5y = 6

y = 6/5

Step 3: Substitute y in x = 5 − y

x = 5 − 6/5

x = 19/5

Solution: x = 19/5, y = 6/5

(ii) 3x + 4y = 10 and 2x − 2y = 2

Step 1: Divide second equation by 2

x − y = 1

x = 1 + y

Step 2: Substitute in first equation

3(1 + y) + 4y = 10

3 + 3y + 4y = 10

7y = 7

y = 1

Step 3: x = 1 + 1 = 2

Solution: x = 2, y = 1

(iii) 3x − 5y − 4 = 0 and 9x = 2y + 7

Step 1: Rewrite equations

3x − 5y = 4

9x − 2y = 7

Step 2: Multiply first equation by 3

9x − 15y = 12

Step 3: Subtract second equation

(9x − 15y) − (9x − 2y) = 12 − 7

−13y = 5

y = −5/13

Step 4: Substitute in 3x − 5y = 4

3x − 5(−5/13) = 4

3x + 25/13 = 4

3x = 27/13

x = 9/13

Solution: x = 9/13, y = −5/13

(iv) x/2 + 2y/3 = −1 and x − y/3 = 3

Step 1: Multiply first equation by 6

3x + 4y = −6

Step 2: Multiply second equation by 3

3x − y = 9

Step 3: Subtract equations

(3x + 4y) − (3x − y) = −6 − 9

5y = −15

y = −3

Step 4: Substitute in 3x − y = 9

3x − (−3) = 9

3x + 3 = 9

3x = 6

x = 2

Solution: x = 2, y = −3

2. Word Problems

(i) Fraction Problem

Let numerator = x and denominator = y

(x + 1)/(y − 1) = 1

x + 1 = y − 1

x − y = −2

x/(y + 1) = 1/2

2x = y + 1

2x − y = 1

Subtract equations:

(2x − y) − (x − y) = 1 − (−2)

x = 3

3 − y = −2

y = 5

Required Fraction = 3/5

(ii) Ages Problem

Let Nuri = x and Sonu = y

x − 5 = 3(y − 5)

x − 3y = −10

x + 10 = 2(y + 10)

x − 2y = 10

Subtract equations:

(x − 3y) − (x − 2y) = −10 − 10

−y = −20

y = 20

x − 2(20) = 10

x = 50

Nuri = 50 years, Sonu = 20 years

(iii) Two-Digit Number

Let tens digit = x and units digit = y

x + y = 9

9(10x + y) = 2(10y + x)

90x + 9y = 20y + 2x

88x − 11y = 0

8x − y = 0

y = 8x

Substitute in x + y = 9

x + 8x = 9

9x = 9

x = 1

y = 8

Required Number = 18