Explanation:
Let x be the initial charge and y be the charge per mile driven.
We can write a system of equations:
[tex]\begin{gathered} 11=x+4y \\ 23=x+10y \end{gathered}[/tex]Using the elimination method we can find y easily:
[tex]\begin{gathered} 11=x+4y \\ - \\ \frac{23=x+10y}{-12=-6y} \end{gathered}[/tex]Solving for y:
[tex]\begin{gathered} 6y=12 \\ y=\frac{12}{6}=2 \end{gathered}[/tex]y = 2 is the charge per mile driven. Replacing this into any of the equations we can find the initial charge x:
[tex]\begin{gathered} 11=x+4\cdot2 \\ 11=x+8 \\ x=11-8=3 \end{gathered}[/tex]Answer:
The inital charge is $3