Question

Apples cost x cents each and oranges cost (x + 2) cents each. Dylan spends $3.23 on apples and $3.23 on oranges. The total of the number of apples and the number of oranges Dylan buys are 36. Write an equation in x.

Answer 1

Let's denote the number of apples Dylan buys as \( a \) and the number of oranges as \( o \).

From the problem, we know that:

1. The cost of apples is \( x \) cents each, so the total cost for apples is \( a \cdot x \) cents.
2. The cost of oranges is \( x + 2 \) cents each, so the total cost for oranges is \( o \cdot (x + 2) \) cents.
3. Dylan spends a total of $3.23 on apples and $3.23 on oranges. In cents, $3.23 is equal to 323 cents, so:

\[
a \cdot x = 323 \quad \text{(1)}
\]
\[
o \cdot (x + 2) = 323 \quad \text{(2)}
\]

4. The total number of apples and oranges is 36:

\[
a + o = 36 \quad \text{(3)}
\]

Now, we can express \( o \) in terms of \( a \) using equation (3):

\[
o = 36 - a
\]

Next, we can substitute this expression for \( o \) into equation (2):

\[
(36 - a)(x + 2) = 323
\]

Now, we can distribute and rearrange this equation:

\[
36(x + 2) - a(x + 2) = 323
\]

Expanding gives:

\[
36x + 72 - ax - 2a = 323
\]

Rearranging to isolate terms in \( a \):

\[
36x + 72 - 323 = ax + 2a
\]

\[
36x - 251 = ax + 2a
\]

Next, replace \( a \) with \( \frac{323}{x} \) from equation (1):

\[
36x - 251 = \left(\frac{323}{x}\right)x + 2\left(\frac{323}{x}\right)
\]

Simplifying the right side yields:

\[
36x - 251 = 323 + \frac{646}{x}
\]

To eliminate the fraction, multiply through by \( x \):

\[
x(36x - 251) = 323x + 646
\]

Rearranging gives us the equation in terms of \( x \):

\[
36x^2 - 251x - 323x - 646 = 0
\]

Combining like terms results in:

\[
36x^2 - 574x - 646 = 0
\]

This is the quadratic equation in \( x \).