We are given the following information.
Person 1 rode the subway three times at the peak fare price and five times at the off-peak fare price for a total cost of $19.50.
Person 2 rode two times at the peak fare price and four times at the off-peak fare price for a total cost of $14.50.
We are asked to find the peak fare price.
Let x represent the peak fare price.
Let y represent the off-peak fare price.
Then we can set up the following two equations for persona 1 and person 2.
[tex]\begin{gathered} 3x+5y=19.50\;\;eq.1 \\ 2x+4y=14.50\;\;eq.2 \end{gathered}[/tex]Separate out the y variable in eq.1
[tex]\begin{gathered} 3x+5y=19.50 \\ 5y=19.50-3x \\ y=\frac{19.50-3x}{5} \end{gathered}[/tex]Now, substitute it into eq.2
[tex]\begin{gathered} 2x+4y=14.50 \\ 2x+4(\frac{19.50-3x}{5})=14.50 \\ 2x+\frac{78-12x}{5}=14.50 \\ \frac{5\cdot2x+78-12x}{5}=14.50 \\ \frac{10x+78-12x}{5}=14.50 \\ 10x-12x+78=14.50\cdot5 \\ -2x+78=72.5 \\ -2x=72.5-78 \\ -2x=-5.5 \\ 2x=5.5 \\ x=\frac{5.5}{2} \\ x=\$2.75 \end{gathered}[/tex]Therefore, the peak fare price is $2.75
