Solving Systems of Equations – Part 1
Now that we have done significant work on solving and graphing linear equations, the next step is to consider
what happens when a situation requires more than one equation to find the solution. For example, let’s say you
are putting on an event and as a promotion idea you want to give away prizes to the first 50 people that arrive.
You decide to call up a design company and order t-shirts and hats with your event name and logo on them.
Suppose you get a quote that says hats will cost $2.00 and t-shirts will cost $5.00 to produce. Your budget to buy
these prizes is $130. How many of each should you order so you can most efficiently use up your budget?
This requires the use of a new type of process that builds on the previous work you have done with writing, solving,
and graphing equations. We are now working with what is called a system of equations.
Definition: Any time you use more than one equation to describe a set of variables we call this a
system of equations or simultaneous equations. Solving a system of equations means that you are
looking for all values that can work for a number of equations at the same time.
You can have systems of equations that involve all types of patterns and graphs, but we are going to start by
looking at the most basic: systems of linear equations.
For the previous situation we can come up with two variables for which we are trying to find values: hats (h) and
t-shirts (t). We know two things about our situation: the total cost and the total number of items given away. This
means we will have two equations relating our variables t and h.
Using this we can create the system of equations:
h  t  50 and 2h  5t  130
Let’s start by looking at some values. What if we decide to split it down the middle and buy 25 hats and 25 t-shirts.
Would that come out to a price of $130?
if h  25 and
t  25
h  t  50

25  25  50
2h  5t  130

2(25)  5(25)  50  125  175
Even though we have met the requirements of the first equation (adding up to 50 items) we did not meet the
second requirement of spending exactly $130 which tells us that the value (25, 25) does not work. The formal way
to say this is that the point (25, 25) is not a solution to this system.
So how do we find the solution then? Do we have to just guess and check until we get it right?
You could guess and check if the answer is simple but if it’s not, there are faster ways to find the solution.
There are 3 key methods we will look at in this tutorial:
1. Solving by graphing
2. Solving by substitution
3. Solving by linear combinations
We will start with one we already know how to do: solving by graphing.
We have two equations that we can graph
h  t  50 and 2h  5t  130
But graphs don’t have an h-axis or a t-axis. How do I know which one is x and which one is y?
It doesn’t matter. You can graph it either way with the same result, just remember in the end which one you called
x and which you called y. For the rest of this problem, let’s call the hats x and t-shirts y. So we have:
x  y  50 and 2 x  5 y  130
Notice how these are in standard form (as opposed to slope-intercept or point-slope form) so if you want to graph
them in this format the easiest way is by plotting the intercepts and connecting the dots. If you have forgotten
how to do this, go back and look at graphing linear equations again.
If you want to put them in your graphing calculator then you might have to put them into slope-intercept (or y = )
form first (depending on the brand of calculator you have).
However you choose to do it, you get the following graph:
T-shirts
20
20
Hats
______
x  y  50
______
2 x  5 y  130
OK, so we have a graph…how does that find the number of hats and t-shirts?
Remember, the way to solve a system of equations is to find the points that work for all the graphs at the same
time. If we look at our graph, how many points satisfy both lines at the same time?
Well, to be on both lines at the same time means that the lines would have to cross at a single point which it turns
out they do. This point is our solution!
If we look at our graph this would be the value (40, 10) meaning that 40 hats and 10 t-shirts would work for both
equations.
Let’s check and see if it actually works:
if h  40 and
t  10
h  t  50

40 10  50
2h  5t  130

2(40)  5(10)  80  50  130
This works! And we know it is our only possible solution because looking at our graph we see there is only one
place where the two lines intersect.
This same technique works with every system of equations:
If you graph a system of equations on the same set of axes, the solutions will appear at the points where the
graphs intersect each other.
In this case, we had two straight lines so they only crossed at one point, leaving us with only one solution. If you
had 3 lines they would all have to go through the same point in order for it to be a solution. Here are some
examples:
3 equations with no common solution
Not Solutions
(don’t hit all 3 at the same time)
3 equations with one solution
Solution
The previous example brings us to an important point: some systems of equations might not have any solutions.
What if you had to find the solution to the following system:
y  2 x  4 and y  2 x  9
Before we even graph this we can tell what is going to happen. Look at the slope for these equations….they are
both 2! This means that when we graph this system we are going to have two parallel lines meaning there will
never be a point where they cross and therefore…no solution!
What about the following system?
2 x  3 y  10 and 6 x  9 y  30
Those look like they could potentially have an intersection. But wait…aren’t these the same equation? If we
multiply the first one by 3 we get the second one meaning these are going to be the exact same line.
You can’t exactly have an intersection when there is only one line. However, even though they are the same
equation we have to act like they are two separate lines, so in this case we have an infinite number of solutions
since they will always technically be crossing one another.
And so we have quite a variety of possibilities when it comes to finding solutions. We might have a finite number
of intersections, no intersections at all, or an infinite number of points. It all depends on the equations we are
given.
This brings us to some important vocabulary terms:
1. When a system of linear equations never cross at a single common point we call this system inconsistent.
Inconsistent systems have no solutions. If a linear system has at least one solution, it is called consistent.
2. When the equations in a linear system end up being multiples of each other, they can end up having an
infinite number of solutions. We call these systems dependent. On the other hand, if the system has a
finite number of solutions with equations that are all unique, it is said to be independent.
Here are some examples:
3x  7 y  12  independent
 & consistent
10 x  8 y  10
5 x  y  1  dependent
 & consistent
10 x  2 y  2 
4 x  9 y  3 
 inconsistent
12 x  27 y  4 
Let’s try one more example:
After a successful afternoon of selling ice cream one summer day, you are counting up your earnings and checking
over your inventory. Your cash register kept track of the number purchases throughout the day which came to 247
total items sold. It also kept a running total of sales which came out to be $366.25. However, it does not count
what items were sold, just the prices. If you were selling Orange Creamsicles for $1.25 and Chocolate Bars for
$1.75, can you figure out how many of each you sold without going back and looking at every individual receipt?
We start by writing our equations. We have two variables: creamsicles and chocolate bars. Let’s call them x and y
respectively. We also have two items of information again: the total sales and the total number of items. This will
give us the following two equations:
1.25 x  1.75 y  366.25
and
x  y  247
First, let’s check to see if these are going to be independent and consistent (meaning they are actually going to
give us a meaningful solution).
Is the slope the same on both?
1.25
1

Nope, so they are consistent.
1.75
1
Is the first one just a multiple of the second one? Nope, so they are independent.
That means they should be end up with a finite number of solutions. Since they are both linear, we know they will
only cross at one point, so we should only have one solution where x is the number of creamsicles and y is the
number of chocolate bars.
Let’s graph it.
Solution
(132, 115)
Chocolate
Bars
We can see there is a solution of 132 creamsicles and 115
chocolate bars but it would be hard to tell exactly where it
was if it were not labeled already. This is where
technology comes into play. If you are going to solve
accurately by graphing, you are going to have to use a
graphing calculator or computer to find the intersection.
Otherwise, this is where graphing hits its limit for
usefulness. To solve problems like this more precisely we
are going to have to use some algebraic methods, namely
the two that we mentioned above:
substitution and linear combinations.
Part 2 will cover these. Happy graphing!
Orange Creamsicles
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