Question

Find two numbers such that the sum of thrice the first and the second is 142 and four times the first exeeds the second by 138

Answer 1

Let us assume, the two numbers are x and y.

Given:

3x + y = 142 ------------1

And also given:

4x – y = 138

y = 4x – 138 --------------2

Substitute the value of y from eqn 2 in eqn 1

3x + 4x – 138 = 142

7x - 138 = 142

7x = 142 + 138

7x = 280

x = 40

Therefore, y = (4 * 40) – 138 = 22

The two numbers are 40 and 22

$\bf{King\:Arrived}$

Answer 2

Answer :

›»› The two numbers are 40 and 22.

Given :

• The sum of thrice the first and the second is 142 and four times the first exeeds the second by 138.

To Find :

• The two numbers = ?

Solution :

Let us assume that, the first number is "x" and second number is "y" respectively.

As it is given that, the sum of thrice the first and the second is 142.

→ 3x + y = 142 ......(1)

As it is also given that, four times the first exeeds the second by 138.

→ 4x - y = 138 ......(2)

Our equations are,

• 3x + y = 142 ......(1)
• 4x - y = 138 ......(2)

Adding equation (1) and equation (2)

→ 3x + y + 4x - y = 142 + 138

→ 3x + 4x + y - y = 142 + 138

→ 7x + y - y = 142 + 138

→ 7x = 142 + 138

→ 7x = 280

→ x = 280/7

→ x = 40

Substitute the value of x in equation (1),

→ 3x + y = 142

→ 3 * 40 + y = 142

→ 120 + y = 142

→ y = 142 - 120

→ y = 22

Hence, the two numbers are 40 and 22.