Answer to Question #116038 in Algebra for AlexB

Question

The University of Maryland and the Howard University have a combined enrollment of
48,142 students. Eight times the enrollment of Howard university is 3736 more than two times
the enrollment of the University of Maryland. What is the enrollment of each university?

Answer 1

Let's assume that the University of Maryland enrollment is m and the enrollment of Howard university is h. Now we can write equations system:

$\begin{cases} m+h=48,142 \\ 8h - 2m = 3,736 \end{cases}$

Let's solve the system

$\begin{cases} m=48,142 - h \\ 8h - 2m = 3,736 \end{cases}$

$\begin{cases} m=48,142 - h \\ 8h - 2*(48,142-h) = 3,736 \end{cases}$

$\begin{cases} m=48,142 - h \\ 8h - 96,284 + 2h = 3,736 \end{cases}$

$\begin{cases} m=48,142 - h \\ 10h = 100,020 \end{cases}$

$\begin{cases} m=48,142 - h \\ h = 10,002 \end{cases}$

$\begin{cases} m=48,142 - 10,002=38,140 \\ h = 10,002 \end{cases}$

Answer: University of Maryland enrollment is 38,140 students and enrollment of Howard University is 10,002 students