Solving Systems of Equations
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1. Solve the following system of
equations. Interpret the solution.
There are many possible ways to
proceed in the solution process to this problem. All will give the same solution and all involve
eliminating one of the variables and getting down to a system of two equations
in two unknowns which you can then solve.
The first solution method involves
solving one of the three original equations for one of the variables. Substitute this into the other two
equations. This will yield two equations
in two unknowns that can be solve fairly quickly.
For this problem we’ll solve the
second equation for x to get.
Plugging this into the first and
third equation gives the following system of two equations.
Or, upon simplification
Multiply the second equation by -2
From this we see that . Plugging this into either of the above two
equations yields . Finally, plugging both of these answers into yields .
In the second method we add
multiples of two equations together in such a way to eliminate one of the
variables. We’ll do it using two
different sets of equations eliminating the same variable in both. This will give a system of two equations in
two unknowns which we can solve.
So we’ll start by noticing that if
we multiply the second equation by -2 and add it to the first equation we get.
Next multiply the second equation
by -5 and add it to the third equation.
This gives the following system of
We can now solve this by
multiplying the second by -2 and adding
From this we get that ,
the same as the first solution method.
Plug this into either of the two equations involving only y and z and we’ll get that . Finally plug these into any of the original
three equations and we’ll get .
You can use either of the two
solution methods. In this case both
methods involved the same basic level of work.
In other cases on may be significantly easier than the other. You’ll need to evaluate each system as you
get it to determine which method will work the best.
Recall that one interpretation of
the solution to a system of equations is that the solution(s) are the
location(s) where the curves or surfaces (as in this case) intersect. So, the three equations in this system are
the equations of planes in 3D space (you’ll learn this in Calculus II if you
don’t already know this). So, from our
solution we know that the three planes will intersect at the point
(0,-1,2). Below is a graph of the three
In this graph the red plane is the
graph of ,
the green plane is the graph of and the blue plane is the graph of
. You can see from this figure that the three
planes do appear to intersect at a single point. It is a somewhat hard to see what the exact
coordinates of this point. However, if
we could zoom in and get a better graph we would see that the coordinates of
this point are .
2. Determine where the following two
In this case we’re looking for
where the circle and parabola intersect.
Here’s a quick graph to convince ourselves that they will in fact. This is not a bad idea to do with this kind
of system. It is completely possible
that the two curves don’t cross and we would spend time trying to find a
solution that doesn’t exist! If
intersection points do exist the graph will also tell us how many we can expect
So, now that we know they cross
let’s proceed with finding the TWO intersection points. There are several ways to proceed at this
point. One way would be to substitute
the second equation into the first as follows then solve for x.
While, this may be the way that
first comes to mind it’s probably more work than is required. Instead of doing it this way, let’s rewrite
the second equation as follows
Now, substitute this into the
This yields two values of y: and . From the graph it’s clear (I hope….) that no
intersection points will occur for . However, let’s suppose that we didn’t have
the graph and proceed without this knowledge.
So, we will need to substitute each y
value into one of the equations (I’ll use ) and solve for x.
So, the two intersection points
are (-2,3) and (2,3). Don’t get used to
these being “nice” answers, most of the time the solutions will be fractions
So, in this case we get complex
solutions. This means that while and are solutions to the system they do not
correspond to intersection points.
On occasion you will want the
complex solutions and on occasion you won’t want the complex solutions. You can usually tell from the problem
statement or the type of problem that you are working if you need to include
the complex solutions or not. In this
case we were after where the two curves intersected which implies that we are
after only the real solutions.
3. Graph the following two curves and
determine where they intersect.