`r=2.24+0.06t`
`p=2.89+0.10x`
In the equation above, r and p represent the price per gallons, in dollars, of regular and premium grades of gasolin, respectively, x months after January 1 of last year. What was the cost per gallon, in dollars, of premium gasoline for the month in which the per gallons price of premium exceeded the per gallon price of regular by $0.93?

Answer

To solve the problem step by step, we will follow the given equations for the prices of regular and premium gasoline and find the month when the price of premium gasoline exceeds the price of regular gasoline by $0.93.

### Step 1: Write down the equations for regular and premium gasoline prices.
- Regular gasoline price: \( r = 2.24 + 0.06t \)
- Premium gasoline price: \( p = 2.89 + 0.10x \)

### Step 2: Set up the equation for the condition given in the problem.
We need to find when the price of premium gasoline exceeds the price of regular gasoline by $0.93. This can be expressed as:
\[ p = r + 0.93 \]

### Step 3: Substitute the expressions for \( p \) and \( r \) into the equation.
Substituting the equations we have:
\[ 2.89 + 0.10x = (2.24 + 0.06x) + 0.93 \]

### Step 4: Simplify the equation.
Now, simplify the right side:
\[ 2.89 + 0.10x = 2.24 + 0.06x + 0.93 \]
\[ 2.89 + 0.10x = 3.17 + 0.06x \]

### Step 5: Rearrange the equation to isolate \( x \).
Subtract \( 0.06x \) from both sides:
\[ 2.89 + 0.10x - 0.06x = 3.17 \]
\[ 2.89 + 0.04x = 3.17 \]

Now, subtract \( 2.89 \) from both sides:
\[ 0.04x = 3.17 - 2.89 \]
\[ 0.04x = 0.28 \]

### Step 6: Solve for \( x \).
Now, divide both sides by \( 0.04 \):
\[ x = \frac{0.28}{0.04} \]
\[ x = 7 \]

### Step 7: Find the price of premium gasoline after 7 months.
Now that we have \( x = 7 \), we can find the price of premium gasoline:
\[ p = 2.89 + 0.10(7) \]
\[ p = 2.89 + 0.70 \]
\[ p = 3.59 \]

### Final Answer:
The cost per gallon of premium gasoline for the month in which the price exceeded the price of regular gasoline by $0.93 is **$3.59**.

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