Write an expression for the total grams of enzyme in terms of $x$ and $y$, then decide whether it should be ≤ or ≥ 100 based on the phrase "a minimum of 100 grams."

Because $x$ and $y$ represent numbers of containers, can they be negative? Use that to determine which inequalities about $x$ and $y$ make sense.

Once you know the correct inequality signs for the cost, the enzyme amount, and the possible values of $x$ and $y$, look for the choice whose inequalities all match those conditions.

In Desmos, enter the equation for the budget boundary line, for example 40x + 60y = 1200. Then decide which side of this line represents spending no more than $1,200. Try shading using an inequality (40x + 60y <= 1200) and see that this corresponds to the region where total cost does not exceed the budget.

Enter the equation 3x + 5y = 100. This line separates combinations that give exactly 100 grams of enzyme from those that give more or less. Shade with an inequality like 3x + 5y >= 100 and notice that the shaded side represents combinations with at least 100 grams of enzyme.

Add x >= 0 and y >= 0 to Desmos. The overlapping shaded area where all inequalities are satisfied represents the realistic combinations of containers. Compare the inequalities you used to the answer choices and select the option that matches them.

Each container of X costs $40 and each container of Y costs $60.

So the total cost is $40x + 60y$.

The laboratory's budget is at most $1,200, which means the total cost cannot be more than $1,200. In math, “at most” translates to a less than or equal to inequality, so this gives an inequality of the form $40x + 60y \le 1{,}200$.

Each container of X has 3 grams of enzyme, and each container of Y has 5 grams.

So the total grams of enzyme from all containers is $3x + 5y$.

The lab needs a minimum of 100 grams of the enzyme. “Minimum of 100” means at least 100, which we write as a greater than or equal to inequality: $3x + 5y \ge 100$.

The problem says $x$ is the number of containers of solution X and $y$ is the number of containers of solution Y.

A number of containers cannot be negative, so both $x$ and $y$ must be greater than or equal to 0. That gives the constraints $x \ge 0$ and $y \ge 0$.

Putting all of the conditions together, we get the full system:

• Budget: $40x + 60y \le 1{,}200$ (at most $1,200 dollars)
• Enzyme: $3x + 5y \ge 100$ (at least 100 grams)
• Nonnegative containers: $x \ge 0$, $y \ge 0$

Looking at the answer choices, this system exactly matches choice D: $40x + 60y \le 1{,}200$, $3x + 5y \ge 100$, $x \ge 0$, $y \ge 0$. This is the correct answer.