Solving Systems of Linear Equations
Once you're given a system of linear equations,
how do you find solutions? As for most problems, there are several methods;
we choose the one that seems most appropriate for the system at hand,
and if that doesn't work out, we go on to try another one.
Substitution Method This system
is most useful for systems of 2 equations in 2
unknowns. The essential idea here is that we solve one of the
equations for one of the unknowns, and then substitute the result into
the other equation.
Example Consider the system
Solve the second equation for x;
this gives x = 5 - y.
Then substitute this instead of x
into the first equation and solve the resulting equation for y:
2 ( 5 - y) + 3y = 5
10 - 2y + 3y = 5
10 + y = 5
y = -5
Going back to the solution for x
in the previous equation, we now see that x
= 5 - (-5) = 10.
Thus there is a unique solution to this equation and it is (x,
y) = (10, -5).
Facts
-
It does not matter which equation we choose first and which second. Just
choose the most convenient one first!
-
It does not matter which unknown we choose first and which second. Again,
just choose the most convenient one first.
Elimination Method This method is useful
for any number of equations in any number of unknowns. It essentially consists
of eliminating the variables from the equations one by one, until the system
looks like an upside-down staircase.
Example Consider the following system
of 3 equations in 3 unknowns:
x +
y = 2
2x + 3y + z = 4
x + 2y + 2z = 6 |
Our goal is to transform this system into an
equivalent system from which it is easy to find the solutions. We now do this
step by step.
-
Subtract 2*(Row1) from Row2 and place the result in the second row; subtract
Row1 from Row2 and place in the third row. Leave Row1 as is.
x + y
= 2
y + z = 0
y + 2z = 4 |
-
Subtract Row2 from Row3, and place the result in row3. Leave Row1 and Row2
as they are.
x + y
= 2
y + z = 0
z = 4 |
From the last form of the system we can deduce the following unique
solution to the system:
z = 4, y = -4, and
x = 2-(-4) = 6
Equivalently, we say that the unique solution to this system is (x,
y, z) = (6, -4, 4).
Comments
-
We have placed the variables in columns. This makes the system easier to
work with now, and prepares us for work with matrices in coming sections.
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Not all the systems are so nice and neat; after going through this process
you might end up with fewer equations than variables, or equations that
contradict each other. In the first case you have a system with infinitely
many solutions ; in the second, the system is inconsistent
and so has no solutions at all. Go to these pages for a discussion of these
situations and the solutions that arise through them.
Exercises
In the following you will receive systems of equations in 2 unknowns
and in 3 unknowns. In each of them,
-
Work through the substitution method
or the elimination method on paper.
-
Decide whether the system has a solution, and if so whether it is unique
or there are infinitely many solutions; click on the appropriate button.
-
Now follow the steps in the window that comes up for each situation.
You may have questions about the results of your computations; if so click
(FAQ-button) to get help.
Exercises
with systems of (up to) 2 equations in 2 unknowns:
Exercises with systems of (up to) 3
equations in 3 unknowns:
Frequently Asked Questions