
Patrick B. answered 05/29/20
Math and computer tutor/teacher
x = 1st class
y = business class
z = coach class
X+y+z = 173
z = 3(x+y)+9 = 3x+3y+9
675x + 590y + 480z = 88910
x + y + (3x+3y+9) = 173
675x + 590y + 480(3x+3y+9) = 88910
the first equation becomes:
4x + 4y = 164
x + y = 41 <--- first equation
the second equation becomes:
675x + 590y + 1440x + 1440y + 4320 = 88910
2115x + 2030y = 84590
substitution:
2115x + 2030 ( 41-x) = 84590
2115x + 83230 - 2030 x = 84590
85x = 1360
x = 16
which forces y = 25
which forces z = 132