Let x be the number of hours spent typing by each student.
Since Mitchell can type at a speed of 3 pages per hour, and has already typed 9 pages, the expression that would model Mitchell's situation is:
[tex]3x+9[/tex]
Similarly, Roxanne can type at a speed of 2 pages per hour, and has already typed 12 pages, the expression that would model Roxanne's situation is:
[tex]2x+12[/tex]
Now, we know that after some time both will have the exact same page count. Let's call this number y. This way, we'll have two equations that describe a system of equations:
[tex]\begin{cases}y=3x+9 \\ y=2x+12\end{cases}[/tex]
We'll solve it by substitution, so we'll solve the first equation for x, substitute in the second equation and then find the value of y, as following:
[tex]\begin{gathered} y=3x+9\rightarrow y-9=3x\rightarrow\frac{1}{3}y-3=x \\ \\ y=2x+12 \\ \\ \rightarrow y=2(\frac{1}{3}y-3)+12\rightarrow y=\frac{2}{3}y-6+12\rightarrow y-\frac{2}{3}y=6 \\ \\ \rightarrow\frac{1}{3}y=6\rightarrow y=18 \\ \end{gathered}[/tex]
This way, we can conlcude that:
[tex]y=18[/tex]
Since we've calculated an expression for x in terms of y , we just have to plug in this value in it to find the value of x:
[tex]\begin{gathered} \frac{1}{3}y-3=x \\ \\ \rightarrow\frac{1}{3}(18)-3=x \\ \\ \rightarrow3=x \end{gathered}[/tex]
Therefore, the solution to our system is:
[tex]\begin{gathered} x=3 \\ y=18 \end{gathered}[/tex]
This way, we can conclude that:
