Below.
Step-by-step explanation:
y=35x,
company B's with the equation
y=25x+20,
and company C's with the equation
y=20x+45.
Graphing the equations helps students visualize the solution.
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Looking at the graph, it appears that the graph that represents the cost for company A is below the other two graphs for any amount of time less that 2 hours, where the graph intersects the line that represents the cost of company B. The graph for company B is below the other two up until 5 hours, where it intersects the graph that represents the cost for company C. In other words, company A is least expensive for 0 to 2 hours, company B is least expensive for 2 to 5 hours, and company C is least expensive for more than five hours.
Students should get in the habit of checking the coordinates of the intersection points algebraically, since it is often the case that the coordinates of the intersection points are not whole numbers (and therefore not easy to read from the graph). This also reinforces the relationship between the algebraic representation of the solution and the graphical representation of the solution.
To find the solution algebraically, consider each pair of equations as a system. Let x be the number of hours it takes to repair the furnace and y be the cost of the repair (without parts).
To find the number of hours for which company A and company B cost the same, consider y=35x and y=25x+20. Substituting for y, we get
35x=25x+20.
The solution to this equation gives the number of hours for which company A and company B cost the same. Solving this equation, we find that the cost of company A and company B is the same for 2 hours of labor. The cost is $70.
To find the number of hours for which company A and company C cost the same, Â consider y=35x and y=20x+45. Substituting for y, we get
35x=20x+45.
Solving this equation, we find that the cost of company A and company C is the same for 3 hours labor, for a total cost of $105.
To find the number of hours for which company B and company C cost the same, consider y=25x+20 and y=20x+45. Substituting for y, we get
25x+20=20x+45.
Solving this equation, we find that the cost of company B and company C is the same for 5 hours labor, for a total cost of $145.
Some additional substitution of values shows that company C is $15 more expensive than A and B at 2 hours. Company B is $10 less than A and C at 3 hours. And company A is $30 more expensive at 5 hours than companies B and C.
As before, we found that company A is the least expensive up to a time of 2 hours, at which point company A and B are the same cost. From 2 hours to 5 hours, company B is the least expensive, and at 5 hours company B and C both cost $145. For more than 5 hours, Company C will be the least expensive.
