Once you have an inequality involving $G$, replace $G$ with the expression $5.5t + 40$ so that the inequality is in terms of $t$.

Treat the compound inequality the same way you would treat a normal equation: first undo the +40 by subtracting 40 from all three parts, then undo the multiplication by 5.5 by dividing all three parts by 5.5. Keep the inequality signs in the same direction.

When you divide 55 and 100 by 5.5, consider rewriting 5.5 as the fraction $11/2$ so you can multiply by its reciprocal $2/11$ to simplify exactly.

In Desmos, enter y = 5.5x + 40. This graph shows the total charge $G$ (on the y-axis) as a function of time $t$ (on the x-axis).

Add two horizontal lines: enter y = 95 and y = 140. These represent the minimum and maximum allowed total charges.

Look at where the line $y = 5.5x + 40$ lies between the horizontal lines $y = 95$ and $y = 140$. Use the intersection points of the line with each horizontal line to read off the smallest and largest $x$-values; that interval in $x$ is the set of $t$ values that satisfy the promotion requirement.

“At least $95 and at most $140” means the total charge $G$ must be between 95 and 140, inclusive.

So write the compound inequality:

You are given $G = 5.5t + 40$. Substitute this into the compound inequality in place of $G$:

Now the inequality is in terms of $t$, which is what we want.

Subtract 40 from all three parts of the compound inequality to move the 40 to the other side:

This simplifies to:

Now divide every part of the inequality by 5.5 to solve for $t$:

Since $5.5 = \frac{11}{2}$, dividing by $5.5$ is the same as multiplying by $\frac{2}{11}$:

• Left side: $55 \div 5.5 = 55 \cdot \frac{2}{11} = 10$
• Right side: $100 \div 5.5 = 100 \cdot \frac{2}{11} = \frac{200}{11}$

So the solution is:

This matches choice D.