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.