Systems of Linear Equations
Ideas	Graphing linear systems Solving linear systems graphically and algebraically Graphing systems of linear inequalities
Objectives	When finished with this section, the student will be able to graph a system of linear equations. When finished with this section, the student will be able to solve a system of linear equations graphically. When finished with this section, the student will be able to solve a system of linear equations algebraically by substitution.
	Approximate completion time: 4 hours

When we have two linear equations or two linear situations, we have a linear system.

There are only three possible types of linear systems.

Let's consider how to find the solution to a linear system and what that solution means.
(You've actually done this a little bit in some previous lessons, so this should be quite familiar to you.)

Example 1 (From a previous lesson)

"Mike's Mowing Co. will mow lawns for a yearly fee of $50 and $12 per mowing. Carl's Cuts charges $120 for a yearly fee and then $6 per mowing." Find and explain the meaning of the solution to this linear system.

Graphically:   The intersection of the lines occurs between 10 and 12 mowings. The cost at that time lies between $175 and $200. The actual intersection point cannot be determined from the graph. One guess could be that the lines intersect about (11, 190) or 11 mowings and $190. This "solution" point means that for about 11 mowings, the cost for either company would be the same at about $190. Furthermore from looking at the graph, the cost for mowing less than 11 times would be cheaper with Mike's Mowing and the cost for mowing more than 11 times would be cheaper with Carl's Cuts.

Algebraically Let m = the # of mowings. Let C = the total cost in $. For Mike's Mowing, C = 12m + 50. For Carl's Cuts, C = 6m + 120. There will be one solution since the two rates are different. This will occur when the costs are the same. The two equations are C = 12m + 50 C = 6m + 120 Since the costs would be the same, 12m + 50 = 6m + 120 Solving for m, we get m = 11 2/3 or 11.6667 Thus, the costs are only the same at 11 and 2/3 mowings. The cost at such at time would be C = 12(11 2/3)+ 50 = 132 + 8 + 50 = 190 Since Mike's Mowings charges more per mowing, their company will cost more than Carl's for 12 or more mowings. Likewise, since Mike's Mowings starts at a lower initial amount, their costs will be less than Carl's for 11 or fewer mowings.

Do you notice that the graphical method does not give an exact amount for the solution?
Do you notice that the algebraic method gives an exact solution?
Which method do you like better?

 

Example 2

"The Hare knows he can run faster than the Tortoise, so he gave the Tortoise a 2 mile head start. The Hare usually runs about 6 miles per hour while the Tortoise can hold a sustained 3 miles per hour all day long." Find and explain the meaning of the solution to this linear system.

Graphically: The intersection of the lines occurs between 0 and 1 hours into the race and about 4 miles into the race. One could approximate the intersection at the point (3/4, 4) which would be about 3/4 of an hour into the race and 4 miles. At that point, the Hare would pass the Tortoise in the race.

Algebraically Let h = the # of hours into the race. Let D = the # of miles into the race. For the Hare, D = 6h. For the Tortoise, D = 3h + 2. There will be one solution since the two rates are different. This will occur when the distances are the same. The two equations are D = 6h D = 3h + 2 Since the distances would be the same, 6h = 3h + 2 Solving for h, we get h = 2/3 Thus, the distances are only the same at 2/3 of an hour. The distance at such at time would be D = 6(2/3) = 4 Since the Tortoise is slower than the Hare, the Tortoise would be passed up at 2/3 of an hour (40 minutes) into the race at the 4 mile mark.

Did you notice that the graph again was not perfectly accurate in its solution?
Did you notice that there was no information about the length of the race?
Who would have won if the race were only 3 miles long? 5 miles? 4 miles?

Now it's your turn to practice.

In the following assignment, you will get to practice solving systems of linear equations.

Click here to download Assignment 7.05