So far, you've dealt with a single equation at a time—or, in the case of absolute value, one after the other. Now it's time to move to multiple equations at one time—to deal with systems of equations. We will focus on two or three (linear) equations at a time…for now.
First solution technique: graphing. The solution to a single equation in two variables (a line) is a point that lies on the graph of the line. Thus, the solution to a system of equations will be a point that lies on each of the graphs. By graphing the equations, we can look for points of intersection—which will be the solutions to the system. A system that has a solution is called an independent system.
Must two lines intersect? Of course not! The lines could be parallel—and you even know how to tell if two lines are parallel! If the graphs of the equations are parallel, then the system has no solution. This system is called inconsistent.
One other thing that can come up when solving systems of equations is that the two equations might actually have the same graph (they're really the same equation, written in two different ways). If this is the case—if you've been given just one unique equation—then there are an infinite number of solutions to the system. This system is called dependent.
[1.] Solve .
The first equation is in blue; the second is in red. Here's the graph:
The solution is .
[2.] Solve .
Same as before:
These lines are parallel; there is no solution.
[3.] Solve .
Graph:
Wait a minute…that looks like just one line! In fact, that is the case—those equations are for the same line. There are an infinite number of solutions.
It wouldn't be algebra class if we didn't try an algebraic solution!
For the substitution method, choose an equation, and solve that equation for one of the variables. Note that this will be an equation, not a number.
Now, take that equation and substitute it into the second equation—which eliminates one of the variables! Solve for the remaining variable, then substitute that value back into one of the original equations—once again, a variable is eliminated! Solve for the value of remaining variable.
For the elimination method, write the equations one on top of the other, so that like terms (and the =) are in columns. Pick a variable—it doesn't really matter which. Find the least common multiple of the coefficients for that variable. Now you must multiply each equation by a number—whatever number will result in your chosen variable being that least common multiple you found. Now subtract within columns…this will eliminate a variable, so that you can solve for the other. Plug that value back into one of the original equations to solve for the remaining variable.
This is easier than it sounds…
Perhaps you'll get a solution—hurrah! However, this is not always the case. Sometimes, you end up eliminating all of the variables, which indicates that there is not a single solution. If, after eliminating all of the variables, you end up with a true statement (like ), then there are an infinite number of solutions. If the variable eliminations result in a false statement (like ), then there is no solution.
[4.] Solve with the substitution method.
Well, that's convenient—the second equation is already solved for y! Plug that into the first equation: . Now that's weird—the variable disappeared, and we are left with a false statement. That means that there is no solution to this system.
[5.] Solve with the substitution method.
Take your pick—solve either equation for either variable. I'll solve the second equation for x: . Substitute that into the first equation: . Solve: . Plug back in: . The solution is . There is nothing wrong with fraction answers!
[6.] Solve: with the elimination method.
First, rearrange into like columns: . Now, subtract within columns: , , and —which gives the equation . This solves to ; the solution to the system is .
[7.] Solve with the elimination method.
First, we've got to line things up.
I can just add these—the coefficients are already the same, and of opposite sign (i.e., no need to subtract). However, when I subtract, I get …no variables, and the statement is true. There are an infinite number of solutions.
We had to solve a single inequality (in two variables) by graphing. The same will be true for multiple inequalities. Because of this, a system of inequalities doesn't get harder if there are more than two or three inequalities—there's just more to graph.
Graph the solution to each of the inequalities in the system. The answer will be the intersection of all shaded regions. There is no solution if not all of the shades come together into an intersection.
This will be difficult to see in a written document; pay attention to the procedure as I demonstrate it in class.
[8.] Solve .
Let's try this: I'll graph the solution to the first inequality only. I've used a shading type much like you might use when working by hand.
Now, I add the solution to the second inequality.
Now—what's been shaded with every type? That region is the solution to the system.
[9.] Solve .
Here are all of the individual solutions put together:
Can you see the solution region?
[10.] Solve .
Here are all of the individual solutions together:
Can you see the solution?
No—because there isn't one! There are some regions that get two of the three shade types, but there is no region that gets all three.
Linear programming is a special technique for solving a particular type of problem. Specifically, there will be some equation which is to be maximized or minimized (the objective function), and a set of constraints on the variables (boundary conditions; a system of inequalities).
The method is quite simple. First, graph the constraints as a system of inequalities. This will (should) result in a convex shaded region with some vertices. The vertices are the candidates for the solution—plug them into the function, and choose the one that does what you want (maximum or minimum).
[11.] Maximize the value of given the following constraints:
First, graph the inequalities:
Now find the coordinates of the vertices:
Check each of these in the objective function :
Thus, the solution is .
[12.] A farmer has 10 acres to plant in wheat and rye. He has to plant at least 7 acres. However, he has only $1200 to spend and each acre of wheat costs $200 to plant and each acre of rye costs $100 to plant. Moreover, the farmer has to get the planting done in 12 hours and it takes an hour to plant an acre of wheat and 2 hours to plant an acre of rye. If the profit is $500 per acre of wheat and $300 per acre of rye how many acres of each should be planted to maximize profits?
Perhaps we should define the variables first—the question asks "how many acres of each," so we'll need variables to represent the acres of wheat and the acres of rye. Let x be the acres of wheat, and let y be the acres of rye. Obviously, and .
This means that represents the total acres planted. We know that the number of acres planted must be at least seven, and not more than ten—thus, .
The farmer spends $200 per acre to plant wheat; thus the amount spent on planting wheat is . The amount spent planting rye is ; so the total amount spent is , and we know that this amount must be between zero and $1200—.
The constraints on time will look a lot like those for money— .
Finally, the function to be maximized is profit—$500 for wheat and $300 for rye makes .
Finally! Graph this system of inequalities:
(note that some of these are actually intersections of two inequalities!)
We just need to know the coordinates of the vertices of this triangle…
Now, plug each of those into the profit equation to see which one gives the maximum profit.
So—four acres of each will give a maximum profit of $3200.
A system of three equations (in three variables) is a bit harder to solve. To graph the system, you would have to be able to graph lines in three dimensions…since I don't have any 3D graph paper, that method is out. Our other methods still work…
Work with two equations at a time. With the first pair, eliminate one of the variables—this will result in an equation with two variables. Now pick a different pair of equations (from the original three) and eliminate the same variable. Now you have two equations in two variables—work on this just like we did earlier in the chapter.
Solve one equation for one variable—it's still an equation; that's okay. Plug this solution into the other two equations—that will eliminate one variable, and leave you with two equations in two variables. Once again, solve this as we did earlier in the chapter.
As was the case in §3-1, there may be none, one, or infinite solutions. The indicators are the same.
[13.] Solve .
Let's use the elimination method for this. After careful consideration, I think I'll eliminate y from the equations. For the first pair of equations, multiply the second equation by 2 and add the equations.
Now for the second pair of equations—I'll multiply the middle equation by two, then subtract the equations.
This leaves us with two equations and two variables. I'll eliminate z (since the numbers are smaller).
Plug back in and find z: .
Plug these into any of the original three to solve for y: .
We should plug these solutions into the other two equations to check our work:
So, the answer is .
[14.] Solve .
Let's try substitution with this one. I'll solve the first equation for x (), then plug that into the other two.
Here comes a substitution trick: I'm going to solve the first equation for ( ), then substitute that into the second equation: .
Now find y: .
Finally, x: . Check this in the other two equations:
The solution is .
[15.] Solve .
I'll use elimination. First up: the first and last equations:
Now, the last two equations:
So now I've got a smaller system to solve:
Oops! The false statement indicates that there is no solution.
[16.] Solve .
I'll try substitution this time. I can solve the first equation for y () and plug that into the second equation (). There isn't a y to plug into in the third equation, but it's already down to x and z…so that leaves me with this system to solve:
No variables, and a true statement—there are an infinite number of solutions.
Page last validated 2010-08-15