Let d and s be the cost of a double and single- occupancy room, respectively. Since a double-occupancy room cost $20 more than a single room, we can write
[tex]d=s+20\ldots(A)[/tex]
On the other hand, we know that 15 double-rooms and 26 single-rooms give $3088, then, we can write
[tex]15d+26s=3088\ldots(B)[/tex]
Solving by substitution method.
In order to solve the above system, we can substitute equation (A) into equation (B) and get
[tex]15(s+20)+26s=3088[/tex]
By distributing the number 15 into the parentheses, we have
[tex]15s+300+26s=3088[/tex]
By collecting similar terms, it yields,
[tex]41s+300=3088[/tex]
Now, by substracting 300 to both sides, we obtain
[tex]41s=2788[/tex]
then, s is given by
[tex]s=\frac{2788}{41}=68[/tex]
In order to find d, we can substitute the above result into equation (A) and get
[tex]\begin{gathered} d=68+20 \\ d=88 \end{gathered}[/tex]
Therefore, the answer is:
[tex]\begin{gathered} \text{ double occupancy room costs: \$88} \\ \text{ single occupancy room costs: \$68} \end{gathered}[/tex]
