Respuesta :
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,
[tex]x+y+z=12[/tex] ...(i)
[tex]2x+3y+4z=37[/tex] ...(ii)
[tex]7x-y=25[/tex] ...(iii)
From (iii), we get
[tex]-y=25-7x[/tex]
[tex]y=-25+7x[/tex]
Put [tex]y=-25+7x[/tex] value in (i).
[tex]x+(-25+7x)+z=12[/tex]
[tex]8x-25+z=12[/tex]
[tex]8x+z=12+25[/tex]
[tex]8x+z=37[/tex] ...(iv)
Put [tex]y=-25+7x[/tex] value in (ii).
[tex]2x+3(-25+7x)+4z=37[/tex]
[tex]2x-75+21x+4z=37[/tex]
[tex]23x+4z=37+75[/tex]
[tex]23x+4z=112[/tex] ...(v)
Now, Multiply equation (iv) by 4 and subtract the result from (v).
[tex]23x+4z-4(8x+z)=112-4(37)[/tex]
[tex]23x+4z-32x-4z=112-148[/tex]
[tex]-9x=-36[/tex]
[tex]x=4[/tex]
Put x=4 in (iv).
[tex]8(4)+z=37[/tex]
[tex]32+z=37[/tex]
[tex]z=37-32[/tex]
[tex]z=5[/tex]
Put x=4 in [tex]y=-25+7x[/tex].
[tex]y=-25+7(4)[/tex]
[tex]y=-25+28[/tex]
[tex]y=3[/tex]
Therefore, the three numbers are 4, 3 and 5 respectively.
