Question

Owen is shopping for a coworker's retirement party. Regular sized boxes of forks contain enough for 16 guests and cost $5, while value-pack boxes contain enough for 25 guests and cost $10. In all, there must be at least enough forks for the 61 guests who plan to attend. Owen only has $60 to spend on cutlery for the party. Write a system of inequalities to describe this scenario and then give two solutions that work for the scenario.

Answer 1

Step-by-step explanation:

hope you can understand

Answer 2

Answer:

3 regular-sized boxes and 2 value-pack boxes

6 regular-sized boxes and 2 value-pack boxes

Step-by-step explanation:

Let x = number of regular-sized boxes of forks

Let y = number of value-pack boxes of forks

Inequality 1

Given:

• Regular sized boxes contain enough forks for 16 guests
• Value-pack boxes contain enough forks for 25 guests
• There must be at least enough forks for the 61 guests

$\implies 16x+ 25y \geq 61$

Inequality 2

Given:

• Regular sized boxes cost $5 each
• Value-pack boxes cost $10 each
• Maximum money to spend = $60

$\implies 5x + 10y \leq 60$

Therefore, the system of inequalities to describe the scenario is:

$\begin{cases}16x + 25y \geq 61\\5x + 10y \leq 60\end{cases}$

To find two solutions that work for the scenario, graph the inequalities.

When graphing inequalities:

< or > : draw a dashed line

≤ or ≥ : draw a solid line

< or ≤ : shade under the line

> or ≥ : shade above the line

Rearrange Inequality 1 to make y the subject:

$\implies 25y \geq 61 - 16x$

$\implies y \geq \dfrac{61}{25} - \dfrac{16}{25}x$

Rearrange Inequality 2 to make y the subject:

$\implies 5x + 10y \leq 60$

$\implies 10y \leq 60-5x$

$\implies y \leq 6-\dfrac{1}{2}x$

Graph the two lines by drawing a solid line for each.

Shade above the line of first inequality and below the line of the second inequality.

Solutions to the inequalities are any points in the shaded area.

However, solutions to the inequalities for this scenario are any points in the shaded area where x and y are positive integers (see second attached image).

Therefore, two solutions that work for the scenario are:

• 3 regular-sized boxes and 2 value-pack boxes
• 6 regular-sized boxes and 2 value-pack boxes

Learn more about inequalities here:

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