Question

Bentley is deciding between two landscaping companies for his place of business. Company A charges $40 per hour and a $175 equipment fee. Company B charges $50 per hour and a $125 equipment fee. Let A represent the amount Company A would charge for t hours of landscaping, and let B represent the amount Company B would charge for t hours of landscaping. Write an equation for each situation, in terms of t, and determine the interval of hours, t, for which Company A is cheaper than Company B.

Answer 1

Let's begin by listing out the information given to us:

time (t) = number of hours used

Company A: charges (A) = $40 per hour, equipment fee (e) = $175

Company B: charges (B) = $50 per hour, equipment fee (e) = $125

Equation 1

$\begin{gathered} A=40t+e \\ A=40t+175----1 \end{gathered}$

Equation 2

$\begin{gathered} B=50t+e \\ B=50t+125----2 \end{gathered}$

Let's proceed by equating both equation 2 & 1 to find the number of hours where they are equal

At what time interval is Company A is cheaper than Company B, we will represent it in inequality form:

$\begin{gathered} 40t+175<50t+125 \\ Subtract,^{\prime}50t^{\prime}\text{from both sides, we have:} \\ 40t-50t+175<50t-50t+125 \\ -10t+175<125 \\ Subtract,^{\prime}175^{\prime}\text{ from both sides, we have:} \\ -10t+175-175<125-175 \\ -10t<-50 \\ \text{Divide both sides by '-10', we have:} \\ \frac{-10t}{-10}<\frac{-50}{-10} \\ t<5 \end{gathered}$

When the number of hours spent is more than 5 hours (>5), company A is cheaper than Company B