Answer:
[tex]\pounds 26.50[/tex]
Step-by-step explanation:
Let
x = cost for one adult
y = cost for one child
z = cost for one senior.
The Reid family:
take 2 adults and 2 children and paid [tex]2x+2y[/tex] that is [tex]\pounds 18[/tex], so
[tex]2x+2y=18[/tex]
The Mghee family:
take 1 senior, 2 adults and 1 child and paid [tex]x+2x+y[/tex] that is [tex]\pounds 18.50[/tex], so
[tex]x+2x+y=18.50[/tex]
The Griffiths family:
take 1 senior and 3 adults and paid [tex]z+3x[/tex] that is [tex]\pounds 19.50[/tex], so
[tex]z+3x=19.50[/tex]
You get the system of three equations:
[tex]\left\{\begin{array}{l}2x+2y=18\\ \\z+2x+y=18.50\\ \\z+3x=19.50\end{array}\right.[/tex]
From the first equation:
[tex]2y=18-2x\\ \\y=9-x[/tex]
From the third equation:
[tex]z=19.50-3x[/tex]
Substitute them into the second equation:
[tex]19.50-3x+2x+9-x=18.50\\ \\-3x+2x-x=18.50-19.50-9\\ \\-2x=-10\\ \\2x=10\\ \\x=5[/tex]
Then
[tex]y=9-5=4\\ \\z=19.50-3\cdot 5=19.50-15=4.50[/tex]
Hence,
the cost for one adult is [tex]\pounds 5[/tex]
the cost for one child is [tex]\pounds 4[/tex]
the cost for one senior is [tex]\pounds 4.50[/tex]
The Linton family takes 1 senior, 2 adults and 3 children and paid
[tex]\pounds 4.50+2\cdot \pounds 5+3\cdot \pounds 4=\pounds 26.50[/tex]
