Mathematics, 26.11.2020 09:00, evazquez

A car salesperson has already sold 30 cars this year. He sells an average of 13 cars per month. Another salesperson has already sold 50 cars this year. She sells an average of 11 cars per month. After how many months will the salespeople have sold the same number of cars?

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answered: jeff7259

After 10 months, both salespersons will sell 160 cars.

Step-by-step explanation:

We are given two different salesperson's statistics of sales. We can use this information to set up a system of equations and solve for our variable.

Let us name the first salesperson Person 1 and the other salesperson Person 2.

Person 1 sells 30 cars this year and sells an average of 13 cars a month.We can represent this with a linear equation (form of y = mx + b) to see the linear relationship at which they sell cars.Because they sell an average of 13 cars each month, this is a recurring amount. Therefore, this is m, or our slope.Because they've already sold 30 cars this year, this is our y-intercept, or b, a.k.a. our starting point of sales.Therefore, we are able to set up the equation for Person 1. This equation is $\text{y = 13x + 30}$ .

Now, for Person 2:

Person 2 sells an average of 11 cars per month and has sold 50 cars already this year.Their sales can be represented the same way - with a linear equation. Therefore, their sales are modeled with the slope-intercept form of an equation of a line (y = mx + b).Considering they have already sold 50 cars, this is where their average sales would start. Therefore, 50 is the y-intercept, or b, of our equation.Since they sell an average of 11 cars per month, they increase their sales by 11 each month. This means that because their sales rise, this is our slope, or m, of our equation.With this information, our equation becomes $\text{y = 11x + 50}$ .

Now, because we have these equations, we can set up a table that will determine the number of months that will elapse before the salespersons will sell the same amount of cars.

For Person 1, the equation will be the increase of the y-value by 13 cars after an initial sale of 30 vehicles. Therefore, we can create a table of values.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 30 \\ \cline{1-2} 1 & 43 \\ \cline{1-2} 2 & 56 \\ \cline{1-2} 3 & 69 \\ \cline{1-2} 4 & 82 \\ \cline{1-2} 5 & 95 \\ \cline{1-2} 6 & 108 \\ \cline{1-2} 7 & 121 \\ \cline{1-2} 8 & 134 \\ \cline{1-2} 9 & 147 \\ \cline{1-2} 10 & 160 \\ \cline{1-2} \end{array}$

For Person 2, we can use the same formatting to create a table. However, the rule changes - we must start at 50 cars being sold and increase it by only 11 cars a month.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 50 \\ \cline{1-2} 1 & 61 \\ \cline{1-2} 2 & 72 \\ \cline{1-2} 3 & 83 \\ \cline{1-2} 4 & 94 \\ \cline{1-2} 5 & 105 \\ \cline{1-2} 6 & 116 \\ \cline{1-2} 7 & 127 \\ \cline{1-2} 8 & 138 \\ \cline{1-2} 9 & 149 \\ \cline{1-2} 10 & 160 \\ \cline{1-2} \end{array}$

Now, we need to see where the y-values are the same in both tables. We can see that we have a value of (10, 160) in both tables, so after 10 months, the salespersons will sell the same amount of cars.

There is an alternative method of solving the problem that is much quicker and will require much less work.

We are given two equations that are both equal to y. Therefore, we can set them equal to each other (dropping the y) and solving for x.

Our value of x will be the amount of months in which the sales are equal.

$\displaystyle{13x+30=11x+50}\\\\2x + 30 = 50\\\\2x = 20\\\\\boxed{x = 10}$

Therefore, after 10 months of sales, the salespersons will have sold the same amount of cars. We can plug this information into one of the equations to see how many cars will be sold at that point.

$y = 13(10) + 30\\\\y = 130 + 30 \\\\y = 160$