Question 1 (Multiple Choice Worth 1 points) \( \begin{array}{l}\text { (05.03 MC) } \\ \text { Harry had } \$ 32 \text {. He spent all the money buying three notebooks for } x \text { dollars each and four packs of index cards for } y \text { dollars each. If Harry had bought five notebooks and five packs o } \\ \text { index cards, he would have run short by } \$ 18 \text {. The following system of equations models this scenario: } \\ \begin{array}{l}3 x+4 y=32 \\ 5 x+5 y=50\end{array} \\ \text { Use the system of equations to solve for } x \text { and } y \text {. }\end{array} \) ?

To solve the system of equations, we have: 1. \( 3x + 4y = 32 \) 2. \( 5x + 5y = 50 \) We can simplify the second equation by dividing it by 5: \[ x + y = 10 \] Now we can express \( y \) from the second equation: \[ y = 10 - x \] Now substitute \( y \) in the first equation: \[ 3x + 4(10 - x) = 32 \] Expanding gives: \[ 3x + 40 - 4x = 32 \] \[ -x + 40 = 32 \] Solving for \( x \): \[ -x = 32 - 40 \] \[ -x = -8 \] \[ x = 8 \] Now substitute \( x \) back to find \( y \): \[ y = 10 - 8 = 2 \] Thus, the solution is \( x = 8 \) and \( y = 2 \). Now, you can envision Harry happily buying his notebooks and index cards, knowing the prices! If you ever find yourself in a similar scenario, remember to eliminate one variable first before plugging it back in—it makes the problem much easier and quicker to solve!