Alex won a gift card worth $96.00 for winning a design competition. She decided to spend her winnings on sketchbooks and paint sets. Each sketchbook (x) costs $12.00, while each paint set (y) costs $15.00. Which of the following combinations of sketchbooks and paint sets can Alex purchase, assuming she spends her entire gift card?

1 answer

To determine the combinations of sketchbooks (x) and paint sets (y) that Alex can purchase with her $96.00 gift card, we can set up the following equation based on the costs of each item:

\[ 12x + 15y = 96 \]

Here,

  • \(x\) represents the number of sketchbooks, and each sketchbook costs $12.00.
  • \(y\) represents the number of paint sets, and each paint set costs $15.00.

To find the combinations of \(x\) and \(y\), we can rearrange the equation into a more manageable form:

  1. Solve the equation for \(y\):

\[ 15y = 96 - 12x \] \[ y = \frac{96 - 12x}{15} \]

  1. Now we need \(y\) to be a non-negative integer. Therefore, \(96 - 12x\) must be a non-negative multiple of 15.

  2. Rearranging gives us the requirement:

\[ 12x \leq 96 \]
Which simplifies to \(x \leq 8\).

  1. Now, we check the values of \(x\) from 0 to 8 and see if \(y\) results in an integer:
  • If \(x = 0\):
    \(y = \frac{96 - 12(0)}{15} = \frac{96}{15} = 6.4\) (not an integer)

  • If \(x = 1\):
    \(y = \frac{96 - 12(1)}{15} = \frac{84}{15} = 5.6\) (not an integer)

  • If \(x = 2\):
    \(y = \frac{96 - 12(2)}{15} = \frac{72}{15} = 4.8\) (not an integer)

  • If \(x = 3\):
    \(y = \frac{96 - 12(3)}{15} = \frac{60}{15} = 4\) (valid combination \(x=3, y=4\))

  • If \(x = 4\):
    \(y = \frac{96 - 12(4)}{15} = \frac{48}{15} = 3.2\) (not an integer)

  • If \(x = 5\):
    \(y = \frac{96 - 12(5)}{15} = \frac{36}{15} = 2.4\) (not an integer)

  • If \(x = 6\):
    \(y = \frac{96 - 12(6)}{15} = \frac{24}{15} = 1.6\) (not an integer)

  • If \(x = 7\):
    \(y = \frac{96 - 12(7)}{15} = \frac{12}{15} = 0.8\) (not an integer)

  • If \(x = 8\):
    \(y = \frac{96 - 12(8)}{15} = \frac{0}{15} = 0\) (valid combination \(x=8, y=0\))

Thus, the valid combinations of sketchbooks and paint sets that Alex can purchase are:

  1. \(x = 3\), \(y = 4\) (3 sketchbooks and 4 paint sets)
  2. \(x = 8\), \(y = 0\) (8 sketchbooks and 0 paint sets)

Therefore, the combinations Alex can purchase are:

  • 3 sketchbooks and 4 paint sets,
  • 8 sketchbooks and no paint sets.