It’s Professor Dave, let’s solve some linear systems.
By now we have gotten pretty good at finding solutions to equations, whether they are linear,
quadratic, or beyond.
But what about when we have a system of two different linear equations.
Can there be a solution to this set?
Can there be a value or set of values that works for both equations?
Usually there will be, so let’s find out some different ways that we can get to this
solution.
First of all, since we are talking about lines, we can graph them.
Say we have Y equals X and Y equals negative X plus four.
If we place both of these lines on the coordinate plane, we can see that they do indeed cross.
If one line represents all the pairs of X and Y values that are solutions to one equation,
and the other line represents all the pairs of X and Y values that are solutions to the
other equation, then it must necessarily be the case that the point where they cross must
be a solution to both equations, because this point is on both lines.
We can find this solution graphically if our graphs are very neat and tidy, or if we are
using a graphing calculator, because as long as the graph is accurate, we can just observe
that the lines cross at the point two two, and therefore X equals two and Y equals two
represents the solution to this system of linear equations.
Most of the time, two lines will cross precisely once, and there will therefore be one unique
solution to the system.
However, other situations include parallel lines, which never cross, so there will be
no solution to the system, and we could also have two different equations that describe
precisely the same line, in which case there are infinitely many solutions, as every point
on the line will be a solution for both equations.
But as we said, we are typically concerned with situations where two lines cross once,
and we want to be able to get that one solution.
When graphing isn’t so easy for one reason or another, there are other methods we can
use to find the solution.
The first technique we will look at is called substitution.
Say we have these two equations, two X plus Y equals four, and X plus two Y equals negative
one.
What we could do is solve for one of these variables in terms of the other.
For example, let’s solve for Y in the first equation.
Subtracting two X from both sides gives us Y equals four minus two X.
Now what we can do is substitute this value of Y for the Y in the other equation.
That means instead of Y, we insert the quantity four minus two X.
Now we have a new equation with only X, which will make it possible to solve for X.
We distribute the two to get eight minus four X, combine like terms, bring the eight to
the other side, and we get negative three X equals negative nine, or X equals three.
That means three is the X coordinate for the point where these lines cross.
To get the Y value, we just plug this X value into either of these equations.
Using the top one, two times three is six, and if six plus Y equals four, we can subtract
six from both sides to get Y equals negative two.
So X equals three and Y equals negative two, or the point three negative two, is the unique
solution to this system.
Substitution isn’t the only technique we can use.
We can use another technique called elimination.
Say we have another set of linear equations, negative three X plus two Y equals eleven,
and five X minus two Y equals five.
Once again, we absolutely can use the substitution method, but it will sometimes be easier to
go another route.
We can combine these equations in such a way so as to get a new one where one of the variables
has disappeared.
To do this, in this particular case, we just add the equations together.
This is valid, because if A equals B and C equals D, then A plus C must equal B plus
D, so we are just exploiting this logical truth to solve a problem.
Negative three X plus five X gives us two X.
Then, two Y plus negative two Y gives us zero.
The Y term goes away.
And eleven plus five gives us sixteen, so dividing by two, X equals eight.
Then we can plug eight into one of these, like the bottom one, we get forty minus two
Y equals five, negative two Y equals negative thirty five, and Y equals thirty five halves.
So this was a pretty fast way to solve this system.
Of course it won’t always be as easy as just adding them together.
Sometimes we will have to subtract.
If both of the Y terms were positive, we would subtract one from the other to get the Y terms
to go away, and we would then have to be careful with our arithmetic, making sure we get the
right sign on all the values in the answer.
Furthermore, we might have to multiply one or both of the equations by some constant
to be able to do elimination.
Say we have eleven X plus two Y equals thirteen, and seven X minus three Y equals nine.
The Y values are pretty close, but we can’t add them together as is and have them cancel.
Instead, we have to realize that the least common multiple of two and three is six, so
we multiply the top equation by three and the bottom equation by two.
That gives us thirty-three X plus six Y equals thirty-nine and fourteen X minus six Y equals
eighteen.
Now we are ready to combine them, because positive six Y and negative six Y will indeed
cancel each other out.
That gives us forty-seven X equals fifty-seven.
We can then solve for X, and plug back into one of the equations to get the Y value.
With lots of multiplying, the solutions can get messy, but that’s just how they go sometimes.
So in summary, we can see that we have three options available to solve for a system with
two equations and two unknowns.
We can either use substitution, by solving for one variable in terms of the other and
plugging that into the other equation, we can use elimination, by manipulating one or
both of the equations in order to combine them in some manner so as to get one variable
to go away, or we can just graph everything and see where the two lines cross.
If they cross once, there is one solution.
If they don’t cross, there are no solutions.
If they are the same line, there are infinitely many solutions.
So why is this useful?
Let’s look at a real world example.
Say that Penelope and Artemis worked a total of seventy hours this week.
However, Penelope worked two hours less than twice the number of hours that Artemis worked.
How many hours did each of them work?
While this example is relatively simple, and if we did some guessing and checking, we would
eventually find the answer, that wouldn’t be doing algebra, and if we learn the algebra,
we can apply it to much more complicated systems where guess and check just isn’t an option.
First, let’s write some equations.
The two of them worked a total of seventy hours, so let’s write P plus A equals seventy.
Then, we know that Penelope worked two hours less than twice what Artemis did.
So P must be equal to two A minus two.
There’s our two equations with two unknowns.
How shall we solve?
In this case, substitution seems pretty easy, as we’ve already solved for P in the bottom
equation.
Let’s put two A minus two in the place of P in the top equation.
That gives us two A minus two plus A, or three A minus two on the left.
We add two to both sides, divide by three, and we get A equals twenty-four.
Then we can use either of these equations to find P. Two times twenty-four is forty-eight,
minus two is forty-six.
So Penelope and Artemis worked twenty-four and forty-six hours this week.
Let’s check comprehension.