SOLUTION: The sum of three numbers is 14. The sum of twice the first​ number, 4 times the second​ number, and 5 times the third number is 62. The difference between 4 times the

Let x, y and z be the 1-st, the 2-nd and the 3-rd numbers, respectively.

Then from the condition you have these three equations in 3 unknowns


 x +  y +  z = 14,        (1)
2x + 4y + 5z = 62,        (2)
4x -  y      = -1.        (3)


From eq(3), express  y = 4x +1  and substitute it into eq(1) and eq(2). You will get


 x + (4x+1)  +  z = 14    (1')
2x + 4(4x+1) + 5z = 62    (2')


Simplify and write it in the standard form


 5x +  z = 13             (1'')
18x + 5z = 58             (2'')


Congratulations !  You reduced the 3x3-system to 2x2-system, which is much easier to solve.


To solve it, apply Elimination method.  For it, multiply eq(1'') by 5  (both sides). You will get


25x + 5z = 65             (1''')
18x + 5z = 58.            (2''')


Now subtract eq((1''')  from eq(2'''). You will get


7x = 65 - 58 = 7  ====>  x = 1.


Then from eq(1'')  z = 13-5x = 13 - 5*1 = 8.


Finally,  y = 14 - x - z = 14 - 1 - 8 = 5.


Answer.  x = 1;  y = 5;  z = 8.