Variables
• x: Amount in Christine's wallet, in dollars
,
• y: Amount in Shen's wallet, in dollars
,
• z: Amount in Boris's wallet, in dollars
Christine, Shen, and Boris have a total of $91, then:
[tex]x+y+z=91\text{ (eq. 1)}[/tex]
Shen has 3 times what Boris has, then:
[tex]y=3z\text{ (eq. 2)}[/tex]
Christine has $9 less than Boris, then:
[tex]x=z-9\text{ (eq. 3)}[/tex]
Substituting equations 2 and 3 into equation 1, and solving for z:
[tex]\begin{gathered} (z-9)+(3z)+z=91 \\ (z+3z+z)-9=91 \\ 5z-9=91 \\ 5z-9+9=91+9 \\ 5z=100 \\ \frac{5z}{5}=\frac{100}{5} \\ z=20 \end{gathered}[/tex]
Substituting z = 20 into equations 2 and 3:
[tex]\begin{gathered} y=3\cdot20=60 \\ x=20-9=11 \end{gathered}[/tex]
Therefore, the final answer is:
• Amount in Christine's wallet: ,$11
,
• Amount in Shen's wallet: ,$60
,
• Amount in Boris's wallet: ,$20
