For example:
2x - y = 1
3x - 2y = 5
Is there a good way to solve these types of equations?
Solutions::
There are several good ways to solve systems of linear equations, and
the best method to use in any given situation is the one that requires
the least amount of work. However, that will depend on the particular
equations that you’re trying to solve. The most popular methods are
inspection, elimination, substitution, intersection, and graphing.
Graphing is usually the hardest, unless you have a graphing calculator
or program handy. Then it can be the easiest, but there is always
room for error in trying to read a point of intersection from a graph.
So graphing is good when you just need an approximate answer, but if
you need a lot of precision, you should probably stick with one of the
other methods.
Inspection is the easiest method, and it's always a good idea to check
to see if you can use it before jumping in with another method. When
can you use inspection? It's easiest when one of the variables has
the same coefficient in both equations, e.g.,
3x + 2y = 12
3x + 5y = 18
To use inspection, you reason this way: "In moving from the first
equation to the second, all that changes is that we're adding 3y. On
the right side, we add 6. So it must be the case that 3y is the same
as 6, which means that y must equal 2."
In fact, all you’re really doing here is using elimination, but not
bothering to do the formal addition or subtraction of equations. If
we write the equations in the opposite order,
3x + 5y = 18
3x + 2y = 12
we can subtract the left sides and the right sides to get
(3x+5y) - (3x+2y) = (18) - (12)
3x - 3x + 5y - 2y = 6
3y = 6
which is what we did the last time, but using insight instead of
equations. The nice thing about elimination is that it continues to
work even when inspection fails, which is to say, when all the
coefficients are different:
3x + 2y = 12
6x + 5y = 26
To make elimination work in a situation like this, you need to
multiply one of the equations by a constant factor so that you end up
in a situation you like better, i.e., with matching coefficients. In
this case, we can multiply the first equation by 2 to get
6x + 4y = 24
6x + 5y = 26
Now we can use inspection or elimination, depending on how much we
trust ourselves to do the work in our heads instead of on paper.
That case was sort of like adding 1/2 and 1/4, where one of the
denominators is already the one you're going to use as a common
denominator. But sometimes you get equations like
3x + 2y = 12
7x + 5y = 26
And this is where elimination starts to seem like more trouble than
it's worth, because you have to multiply both equations by different
constants to get matching coefficients:
7(3x + 2y) = 7(12)
3(7x + 5y) = 3(26)
At this point, other methods start to look pretty good. The remaining
methods, substitution and intersection, both use the same first step,
which is to select one of the equations and solve it for one of the
variables in terms of the other. In the example above, we might
choose the first equation and do this:
3x + 2y = 12
2y = 12 – 3x
y = 6 – (3/2)x
At this point, if we want to use substitution, we go back to the other
equation, locate every occurrence of y, and replace it with the
expression 6 – (3/2)x:
26 = 7x + 5y
= 7x + 5(6 – (3/2)x)
Now we're in familiar territory again: one equation, with one
variable. We can solve this to find the value of x, and plug that into
either equation to find the value of y.
In intersection, we reason a little differently. Remembering that
we're dealing with lines, we think: "Assuming the lines aren't
parallel, they have to intersect somewhere, and wherever that happens,
they have to have the same value of y (or of x)." So we solve both
equations for the same variable, and set the expressions equal to each
other:
3x + 2y = 12 becomes y = (12 – 3x)/2
7x + 5y = 26 becomes y = (26 – 7x)/5
and since it must be true that y=y, it must also be true that
(12 – 3x) /2 = (26 – 7x)/5
And again, we’re in familiar territory.
Personally, I prefer elimination to substitution and intersection,
because I like to work with integers whenever possible. Of course, if
I'm starting out with coefficients that aren't integers, then that
excuse goes out the window, and I'll usually reach for intersection.
(I prefer intersection to substitution because I think the equations
tend to look a little nicer.)
Of course, everyone is different, which is why we have all these
methods in the first place. At the beginning of this message, I said
that the best method is the one that requires the least work. I guess
I’d say that the amount of work is only the second most important
consideration. The best method is the one that you have the most
confidence in. (As they say, you can't make mistakes fast enough to
get a correct answer.) And that's something that you can decide only
after you've had some practice with all the different methods.
I hope this helps!
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