Question

1. lexi is making a flower arrangement in her floral design class. the arrangement requires no more than 25 carnations and roses combined. she needs to use at least 3 more carnations than roses. the graph of this system of inequalities is shown. which is a viable solution to the system of inequalities? -4 roses and 10 carnations 10.5 roses and 10 carnations 5 roses and 15 carnations 10 roses and 5 carnations

Explanation:

Step1: Analyze non - negative constraint

The number of roses and carnations can't be negative. For the option “-4 roses and 10 carnations”, the number of roses is - 4, which is not possible in a real - world context of counting flowers, so we eliminate this option.

Step2: Analyze the number of flowers (non - fractional)

Since we are dealing with the number of roses and carnations, they should be whole numbers (you can't have a fraction of a flower in a flower arrangement). The option “10.5 roses and 10 carnations” has a non - whole number of roses, so we eliminate this option.

Step3: Check the inequality \(c\geq r + 3\) (where \(c\) is the number of carnations and \(r\) is the number of roses)

For the option “10 roses and 5 carnations”: Substitute \(r = 10\) and \(c = 5\) into \(c\geq r+3\). We get \(5\geq10 + 3\), which simplifies to \(5\geq13\). This is false, so this option does not satisfy the inequality.

Step4: Check the inequalities for “5 roses and 15 carnations”

First, check the total number of flowers: \(r + c=5 + 15=20\leq25\), which satisfies the “no more than 25 carnations and roses combined” condition.
Then, check \(c\geq r + 3\): Substitute \(r = 5\) and \(c = 15\) into \(c\geq r + 3\). We get \(15\geq5 + 3\), which simplifies to \(15\geq8\). This is true. Also, the number of roses (\(r = 5\)) and carnations (\(c = 15\)) are non - negative whole numbers.

Answer:

5 roses and 15 carnations