Example from lecture, explained in more detail
Question: For which values of s, t do we have a unique solution, infinitely many solutions, and
no solutions to the system of equations below?
À
x + 2y = 10
Á 3x + (6 + t)y = s.
Solution: To eliminate x, we multiply the first equation through by −3 and add it to the
second equation:
−3 · À
−3x − 6y = −30
Á 3x + (6 + t)y = s
ty = s − 30
Add:
After this step is, we have changed the original system of equations into the following equivalent
system:
À x + 2y = 10
Á
ty = s − 30.
(∗)
To solve for y, we would like to divide by t, but we cannot do that if t = 0. So let’s consider
first the case t = 0.
• Case: t = 0: Our system of equations is then:
À x + 2y = 10
Á
0 = s − 30.
– If s 6= 30, the second equation cannot be true. For instance, if s = 32, we would have
0 = 2, which is false! Thus for t = 0, s 6= 30, the system has no solutions.
– If s = 30, the second equation holds. The system of equations is:
À x + 2y = 10
Á
0 = 0.
The solutions of this system are all the values of x and y that satisfy both equations.
Since 0 = 0 doesn’t depend on x and y, it’s enough to consider the values that satisfy
x + 2y = 10. This is a line consisting of infinitely many points. For instance x = 2 and
y = 4 is one solution; x = 8 and y = 1 is another solution. In fact the solutions to this
equation are
x = 10 − 2r
y = r
where r can be any real number. Thus for t = 0, s = 30, the system has infinitely
many solutions.
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• Case: t 6= 0: Going back to the system (∗), if t 6= 0 we can divide the second equation by t.
Our new system of equations is then
À x + 2y = 10
Á
y = s−30
t .
We can use back-substitution to solve for x, by substituting the value
equation:
= 10
x + 2 s−30
t
x = 10 − 2(s−30)
.
t
s−30
t
for y in the first
Our unique solution is
x = 10 −
y = s−30
t
2(s−30)
t
For instance, if t = 2, s = 32, the solution would be x = 8, y = 1. Thus for t, s any
numbers such that t 6= 0, the system has a unique solution.
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