Question

the sum of two numbers is 36 Three times one e exceeds twice the other by 8 find them​

Answer 1

Gɪᴠᴇɴ :-

The sum of two numbers is 36 Three times one e exceeds twice the other by 8.

ᴛᴏ ғɪɴᴅ :-

• Two numbers

sᴏʟᴜᴛɪᴏɴ :-

Let 1st number be x & second number be y

Then,

Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ Cᴏɴᴅɪᴛɪᴏɴ -1 :-

• ( x + y) = 36. ---(1)

Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ Cᴏɴᴅɪᴛɪᴏɴ -2 :-

• 3x - 2y = 8. --(2)

On multiplying eq (1) with 3 to make x variable equal, we get,

➭ 3{ ( x + y) = 36 }

➭ 3x + 3y = 108

On subtracting (2) from (1) , we get

➭ (3x + 3y) - (3x - 2y) = 108 - 8

➭ 3x - 3x + 3y + 2y = 100

➭ 5y = 100

➭ y = 20

Put y = 20 in (1) , we get,

➭ ( x + y) = 36

➭ x + 20 = 36

➭ x = 36 - 20

➭ x = 16

Hence,

• First number = x = 16

• Second Number = y = 20

Answer 2

GIVEN

The sum of two numbers is 36 Three times one e exceeds twice the other by 8

TO FIND

Find the number

SOLUTION

★ Let the number be x and y

According to the given condition

★ The sum of two numbers is 36

• x + y = 36 ---(i)

★ Three times one e exceeds twice the other

by 8

• 3x = 2y + 8

→ 3x - 2y = 8 ---(ii)

Multiply (i) by 2 and (ii) by 1

• 2x + 2y = 72
• 3x - 2y = 8

Add both the equations

→ (2x + 2y) + (3x - 2y) = 72 + 8

→ 2x + 2y + 3x - 2y = 80

→ 5x = 80

→ x = 80/5 = 16

Putting the value of x in equation (i)

→ x + y = 36

→ 16 + y = 36

→ y = 36 - 16 = 20

Hence,

Required numbers are 16 and 20