At the flower shop, a rose and a lily cost \( \$ 8 \), and a rose and a carnation cost \( \$ 10.50 \). If Cinderella bought 6 roses, 2 lilies, and a carnation for \( \$ 34 \), what is the cost of one lily?

To solve the problem, let's denote the cost of a rose as \( r \), the cost of a lily as \( l \), and the cost of a carnation as \( c \). From the problem, we have the following equations based on the costs: 1. \( r + l = 8 \) (the cost of a rose and a lily) 2. \( r + c = 10.50 \) (the cost of a rose and a carnation) Cinderella's total purchase can be described by the equation: 3. \( 6r + 2l + c = 34 \) Now, let's solve the equations step by step. From equation 1, we can express \( l \) in terms of \( r \): \[ l = 8 - r \] From equation 2, we can express \( c \) in terms of \( r \): \[ c = 10.50 - r \] Now we'll substitute \( l \) and \( c \) back into equation 3: \[ 6r + 2(8 - r) + (10.50 - r) = 34 \] Distributing and combining like terms: \[ 6r + 16 - 2r + 10.50 - r = 34 \] \[ (6r - 2r - r) + 26.50 = 34 \] \[ 3r + 26.50 = 34 \] Next, we isolate \( r \): \[ 3r = 34 - 26.50 \] \[ 3r = 7.50 \] \[ r = 2.50 \] Now that we have the cost of a rose \( r \), we can find \( l \) using the first equation: \[ l = 8 - r = 8 - 2.50 = 5.50 \] Lastly, to confirm the calculation, let's find the cost of the carnation \( c \): \[ c = 10.50 - r = 10.50 - 2.50 = 8 \] So the costs are: - Rose \( r = 2.50 \) - Lily \( l = 5.50 \) - Carnation \( c = 8 \) Thus, the cost of one lily is \( \boxed{5.50} \).