Systems of Equations
Start with video below. Then move down the page to examples and applications.
Key Words to look for:
System
Solution Point
Solution Set
Linear System
Coincident Lines
Graph
Intersection
Nonlinear System
Substitution
Back Substitution
Elimination
Examples
Solve each of these systems.
Prove your answers using a graphical and an algebraic method.
When algebraic is specified the method of solution is your choice. Different systems can take advantage
of different features of the system, but you should always arrive at the same conclusion.
2x 3 y 5
Example 1:
y 2 x 2 5x 3
Start with a graphical approach to get a feeling for the kind of solution
set you might discover.
Before we start, we should know if we are to find 0, 1 or 2 solutions. So
we graph first and solve second as a matter of practice. Some would think this is cheating, but informed
decisions can eliminate a ton of doubt!
When we graph it, we get a quick answer to the problem: NO SOLUTION!
Now on to the algebraic solution. Remember the instructions said prove your answer both graphically and
algebraically.
Usually for nonlinear systems, substitution is about the best way to go when solving algebraically.
Now that we know what to look or we should not be surprised by the algebraic result:
2x 3y 5
The Discriminant of the Quadratic after substitution is
y 2 x2 5x 3 negative. There are no real solutions for the system.
2 x 3(2 x 2 5 x 3) 5
(Lucky for you we have no interest in imaginary solutions
2 x 6 x 2 15 x 9 5 here.)
6 x 2 13 x 14 0 A "No Solution" answer requires more justification than a
numeric response. Stating no solution without a correct
b 2 4ac 132 4(6)(14) justification is absolutely of no value.
b 2 4ac 137
Notice also, the original quadratic equation in the system
does have real zeros. This is irrelevant to answering our
question about solutions for the system.
2 x 3 y 12
Example 2:
y x2 7x 8
2x 3y 5
Again, start with a graphical approach to get a feeling for the kind of
solution set you might discover. y 2 x2 7 x 8
This time we see two real solutions. You can solve for them graphically as 2 x 3 2 x 2 7 x 8 12
mentioned in the video.
6 x 2 23x 29 0
Now on to the algebraic solution. In this case, it looked difficult and turned 23 1225
about to be fairly simple! x 1 or 29
2(6) 6
We did the substitution as before. On evaluating the discriminant, we got a perfect square! The equation has two
rational solutions.
The details are below. Having found each of the y-values, you must then
substitute each of them into one of the equations to find the related x-
values. This is usually simplified by using the lowest degree equation in
the system. Be certain to state all solution points unless there is a
restriction developed by an application.
23 1225
x 1 or 29
2(6) 6
Substitute into 2 x 3 y 5
For x 1 For x 29
6
2(1) 3 y 5 2(29 ) 3 y 5
Getting the x-value Getting the y-value 6
3y 3
29 3 y 5
y 1 3
y 44
9
6
Solutions: 1,1 and 29 , 44
9
2 x 3 y 12
Example 3:
7 x y 8
Again, start with a graphical approach to get a feeling for the kind of
solution set you might discover. This time we see a unique solution
point.
You can solve for it graphically, but the result will be at best a decimal
approximation.
The algebraic solution is as simple as it seems afterthe 2 x 3 y 12 y 7x 8
8
first two examples.
y 7x 8 y 7 36
2 x 3 7 x 8 12 23
23x 24 12 y 68
23
x 36
23
Solution: 36
23
, 68
23
9 x 2 y 18
Example 4: Finally, a nice linear system!
2x 5 y 1
Again, start with a graphical approach to get a feeling for the kind of solution set you might discover. This time
we see a unique solution point. You can solve for it graphically, but the result will be at best a decimal
approximation. Now on to the algebraic solution. This is as simple as it seems after the first two ex maples.
The Solution
9 x 2 y 18 2(9 x 2 y 18)
y 27
2x 5 y 1 9(2 x 5 y 1) 49
18x 4 y 36 2x 5 y 1
18x 45 y 9
49 y 27
2 x 5 27
49
1
x 92
y 27 49
49
The Graph
SolutionPoint: 92
49
, 27
49 Example 5: This is a financial
application.
An investor has two ways to
invest a total of $25,000 . He
must invest all of his money in
any reasonable combination in
accounts yielding 6% and 9% simple interest for one year. He needs to earn at least $2,000. How should he
divide the money to guarantee his $2,000 requirement?
Problems of this type are handled well using systems. We need to assign our variables:
Let x = amount of money invested at 6% and y = amount of money invested at 9%.
The amount earned from each part is calculated using I = Prt with t fixed at one year. Then we get these two
equations and the graphs.
The solution is to put $16,666.67 in the 9% account with the remainder of $8,333.33 in the other account.
The practicalities of the problem include:
The exact value includes thirds of a penny. We don't have a third-of-a-cent coin!
We must round up the highest interest rate to get the the desired minimum.
Then we must make certain the total is still $25,000.
Example 5: This is a business application.
There are two elements of merchandising that often create a natural system of equations. They are supply and
demand. In an ideal world the manufacturer produces exactly as much as his consumers need and vise-versa. In
reality he builds it, which creates a supply. Then they check the price and how badly they need it before they
buy. This creates the demand. The merchandiser needs to find the optimum price for his product called an
equilibrium value where his supply and their demand balance.
The supply equation tells the supplier the price p necessary to make x unit available to the market (the
customers). It is usually a linear function of price with a positive slope. It could be written this way
p b1 k1x for b1, k1 0 .
The demand equation reflects how many units can be sold according to the customers interest in the item at that
price. It is usually a linear function of price with a negative slope. It could be written this way
p b2 k2 x for b2 , k2 0 .
Notice that neither of these equations is a profit, cost or revenue statement. They have different purposes. If the
demand price for an item is below the manufacturer's average cost for that number of items, he is pretty much
out of business since he cannot sell emough of his wares to break even.
Suppose that an item has a demand price calculated by p 191 0.000060 x and a supply equation of
p 115 0.000320 x . Find the equilibrium value and the selling price at equilibrium.
The correct answers are 200,000 items of sales and $179. You should do this yourself to check the answers.