WORD PROBLEMS WITH TWO UNKNOWNS USING LINEAR EQUATIONS

Problem 1 :

The sum of two numbers is 137 and their difference is 43. Find the numbers.

Solution:

Let us consider, first number = x and second number = y

x + y = 137 --- > (1)

x - y = 43 --- > (2)

Adding both the equation, we get

2x = 180

x = 90

By applying x = 90 in (1)

90 + y = 137

y = 137 - 90

y = 47

So, the numbers are 90 and 47.

Problem 2 :

The sum of thrice the first and the second is 142 and four times the first exceeds the second by 138, then the numbers.

Solution:

Let us consider, first number = x and second number = y

3x + y = 142 ---> (1)

4x - y = 138 --- > (2)

Add (1) and (2), we get

7x = 280

x = 40

By applying x = 40 in (1),

3(40) + y = 142

120 + y = 142

y = 142 - 120

y = 22

So, the numbers are 40 and 22.

Problem 3 :

Sum of two numbers is 50 and their difference is 10, then find the numbers.

Solution:

Let us consider, first number = x and second number = y

x + y = 50 --- > (1)

x - y = 10 --- > (2)

Adding both the equation, we get

2x = 60

x = 30

By applying x = 30 in (1)

30 + y = 50

y = 50 - 30

y = 20

So, the numbers are 30 and 20.

Problem 4 :

The sum of twice the first and thrice the second is 92 and four times the first exceeds seven times the second by 2, then find the numbers.

Solution:

Let us consider, first number = x and second number = y

2x + 3y = 92 --- > (1)

4x - 7y = 2 --- > (2)

(1) × 2 ==> 4x + 6y = 184

Subtract (1) from (2)

13y = 182

y = 182/13

y = 14

By applying y = 14 in (1),

2x + 3(14) = 92

2x + 42 = 92

2x = 92 - 42

2x = 50

x = 50/2

x = 25

So, the numbers are 25 and 14.

Problem 5 :

The sum of two numbers is 1000 and the difference between their squares is 256000, then find the numbers.

Solution:

Let the numbers are x and y.

x + y = 1000 --- > (1)

x² - y² = 256000

x² - y² = (x + y) (x - y)

1000 × (x - y) = 256000

x - y = 256 --- > (2)

Adding (1) and (2),

2x = 1256

x = 1256/2

x = 628

By applying x = 628 in (1),

628 + y = 1000

y = 1000 - 628

y = 372

Problem 6 :

The difference between two numbers is 14 and the difference between their squares is 448, then find the numbers.

Solution:

Let us consider, first number = x and second number = y

x - y = 14 --- > (1)

x² - y² = 448

(x + y) (x - y) = 448

(x + y) × 14 = 448

x + y = 448/14

x + y = 32 --- > (2)

Adding (1) and (2)

2x = 46

x = 46/2

x = 23

By applying x = 23 in (1)

23 - y = 14

-y = 14 - 23

y = 9

So, the numbers are 23 and 9.

Problem 7 :

The sum of two natural numbers is 8 and the sum of their reciprocals is 8/15. Find the numbers.

Solution:

Let the number be x and y.

x + y = 8 --- > (1)

1/x + 1/y = 8/15

(y + x) / xy = 8/15

8/xy = 8/15

xy = 15

y = 15/x

By applying y = 15/x in (1)

x + 15/x = 8

x² + 15 = 8x

x² - 8x + 15 = 0

x² - 3x - 5x + 15 = 0

x (x - 3) - 5 (x - 3) = 0

(x - 5) (x - 3) = 0

x = 5 or x = 3

So, the two natural numbers are 5 and 3.