Respuesta :
x= adult passes
y= child passes
z=ride tickets
the first family:
[tex]2x+3y+9z=29\text{ (1)}[/tex]the sencond family
[tex]x+2y+8z=19\text{ (2)}[/tex]third family
[tex]3x+5y+21z=51\text{ (3)}[/tex]now we have the 3 equations, and we can solve x, y and z
for the equation of the second family we have:
[tex]x=-2y-8z+19\text{ (4)}[/tex]reeplace the new equation(4) in (1), we have:
[tex]2(-2y-8z+19)+3y+9z=29[/tex][tex]-4y-16z+38+3y+9z=29[/tex][tex]y+7z=9\text{ (5)}[/tex]reeplace (4) in (3)
[tex]3(-2y-8z+19)+5y+21z=51[/tex][tex]-6y-24z+57+5y+21z=51[/tex][tex]-y-3z=-6[/tex][tex]y+3z=6\text{ (6)}[/tex]with 5 and 6, we have a 2x2 equation
that we can solve easier
solving 5 and 6, we have:
[tex]\begin{gathered} y+7z=9 \\ y=9-7z\text{ (7)} \end{gathered}[/tex]reeplace 7 in 6
[tex]\begin{gathered} 9-7z+3z=6 \\ 4z=3 \\ z=\frac{3}{4}=0.75 \end{gathered}[/tex]now we find y, reeplace z in (7)
[tex]\begin{gathered} y=9-7(0.75) \\ y=9-5.25 \\ y=3.75 \end{gathered}[/tex]and finally we can find x, reeplacing y and z in (4)
[tex]\begin{gathered} x=-2y-8z+19 \\ x=-2(3.75)-8(0.75)+19 \\ x=-7.5-6+19 \\ x=5.5 \end{gathered}[/tex]