Question

The sum of three numbers is 12 . The sum of twice the first​ number, 3 times the second​ number, and 4 times the third number is 37 . The difference between 7 times the first number and the second number is 25. Find the three numbers.

Answer 1

Given:

The sum of three numbers is 12 .

The sum of twice the first​ number, 3 times the second​ number, and 4 times the third number is 37 .

The difference between 7 times the first number and the second number is 25.

To find:

The three number.

Solution:

Let the three numbers are x, y and z respectively.

According to the question,

`x+y+z=12` ...(i)

`2x+3y+4z=37` ...(ii)

`7x-y=25` ...(iii)

From (iii), we get

`-y=25-7x`

`y=-25+7x`

Put
`y=-25+7x` value in (i).

`x+(-25+7x)+z=12`

`8x-25+z=12`

`8x+z=12+25`

`8x+z=37` ...(iv)

Put
`y=-25+7x` value in (ii).

`2x+3(-25+7x)+4z=37`

`2x-75+21x+4z=37`

`23x+4z=37+75`

`23x+4z=112` ...(v)

Now, Multiply equation (iv) by 4 and subtract the result from (v).

`23x+4z-4(8x+z)=112-4(37)`

`23x+4z-32x-4z=112-148`

`-9x=-36`

`x=4`

Put x=4 in (iv).

`8(4)+z=37`

`32+z=37`

`z=37-32`

`z=5`

Put x=4 in
`y=-25+7x`.

`y=-25+7(4)`

`y=-25+28`

`y=3`

Therefore, the three numbers are 4, 3 and 5 respectively.