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
![d=s+20\ldots(A)](https://img.qammunity.org/2023/formulas/mathematics/college/n2tt07cuc8c6s88q05ofuemt9xwvujkt65.png)
On the other hand, we know that 15 double-rooms and 26 single-rooms give $3088, then, we can write
![15d+26s=3088\ldots(B)](https://img.qammunity.org/2023/formulas/mathematics/college/evwk0omyk4vl0bwxp1z9ocq14px8fjdkx3.png)
Solving by substitution method.
In order to solve the above system, we can substitute equation (A) into equation (B) and get
![15(s+20)+26s=3088](https://img.qammunity.org/2023/formulas/mathematics/college/q4l8d7f9yum61c1fj1zu5vkubuivg5kaua.png)
By distributing the number 15 into the parentheses, we have
![15s+300+26s=3088](https://img.qammunity.org/2023/formulas/mathematics/college/1qw6kq6kx3uhkdmvxwgvg1ulrjatjlri1j.png)
By collecting similar terms, it yields,
![41s+300=3088](https://img.qammunity.org/2023/formulas/mathematics/college/mt7ahvz5ancpnzluy2hgazfi9r64e7tm3v.png)
Now, by substracting 300 to both sides, we obtain
![41s=2788](https://img.qammunity.org/2023/formulas/mathematics/college/x0yawsuddbggtw38t39418hcmx1mfuyn3i.png)
then, s is given by
![s=(2788)/(41)=68](https://img.qammunity.org/2023/formulas/mathematics/college/bu4tqen6bb059r788d99q79fa3mw690hbz.png)
In order to find d, we can substitute the above result into equation (A) and get
![\begin{gathered} d=68+20 \\ d=88 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/50pm80tkbb7ytynq6md2avasvbkdrajl1k.png)
Therefore, the answer is:
![\begin{gathered} \text{ double occupancy room costs: \$88} \\ \text{ single occupancy room costs: \$68} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/a94sz868nlfvvl2oev3piyjfdkcij2ugij.png)