Let be:
• r: The price of a grade of regular gasoline.
,
• p: The price of a grade of premium gasoline.
Then, from the word problem, we know that:
![\begin{gathered} p=0.22+r\Rightarrow\text{ Equation 1} \\ 5r+10p=61.30\Rightarrow\text{ Equation 2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/puoc9yn2wh6x7vczbpbfh9q9soq7z8ccn3.png)
Thus, we can write the following equation:
![5r+10(0.22+r)=61.30](https://img.qammunity.org/2023/formulas/mathematics/college/7r9lzyczij86nv4o5ox7j0o85adobfbo3v.png)
Now, we can solve the above equation for r:
![\begin{gathered} 5r+10(0.22+r)=61.30 \\ \text{ Apply the distributive property} \\ 5r+10\cdot0.22+10\cdot r=61.30 \\ 5r+2.2+10r=61.30 \\ \text{ Add similar terms} \\ 2.2+15r=61.30 \\ \text{ Subtract }2.2\text{ from both sides of the equation} \\ 2.2+15r-2.2=61.30-2.2 \\ 15r=59.1 \\ \text{ Divide by 15 from both sides of the equation} \\ (15r)/(15)=(59.1)/(15) \\ r=3.94 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7g0kdmybwtd5nuaxhfw9za4vqojrpr620x.png)
Finally, we replace the value of r in Equation 1, and we solve it for p:
![\begin{gathered} p=0.22+r \\ p=0.22+3.94 \\ p=4.16 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wbf864nq8tl41sonh9ff0tp1htincowgaf.png)
Therefore, the price of a grade of regular gasoline is $3.94, and of a grade of premium gasoline is $4.16.