1. Use elimination:
Example: 2x + 4y
+ z = -4
2x - 4y + 6z = 13
4x - 2y + z = 6
Multiply two equations (if necessary) to
make
the coefficients of x (or y) opposite
each
other.
Choose two equations with which to work. I will work with the
first
two.
Just changing the second sentence to its opposite and adding it to the
first will work here.
That will eliminate x.
2x + 4y + z = -4
-2x + 4y - 6z = -13
Add the two equations:
8y -5z = -17
Do the same procedure again with another pair of
equations, eliminating the same variable. I will
work with the first and last.
-2(2x + 4y + z =
-4)
-4x - 8y - 2z = 8
4x - 2y + z =
6
4x - 2y + z = 6
Add the two
equations:
-10y - z = 14
Now, combine the two sentences in whichx
was
eliminated and eliminate a second variable.
I will eliminate z.
8y -5z = -17
8y -5z = -17
-5( -10y - z = 14
)
50y + 5z = -70
58y = -87
y = -87/58 = -3/2
Substitute y into
one of the equations in two variables and solve for z (in
this case).
I will use the second one.
-10 (-3/2) - z = 14
15 - z = 14
- z = -1
z = 1
Substitute the values found for y and z into one
of the first two equations.
2x + 4(-3/2) + 1 = -4
2x - 6 + 1 = -4
2x -5 = -4
2x = 1
x = 1/2
Solution: (1/2, -3/2, 1)
2. Use matrix equations:
Let A and B
be matrices for which we know the values. X is an
unknown
matrix.
If AX = B, then
x
= A-1B.
Put matrices A and B in a calculator and type A-1×B.
3. Some systems have no solutions. This is what occurs
if
a false statement is reached when
trying to solve the system.
4. A system can have infinitely many solutions. This is
what occurs if a completely true statement
is reached, such as 1 = 1.
The pattern for the solutions can be found.
Example:
x + y - 3z = -1
y - z = 0
-x + 2y = 1
Add the first equation to the third equation: 3y - 3z = 0, so y = z.
Add -3 times the second equation to the equation just obtained: 0 = 0.
Now, the first equation and the
second
equation determine the pattern of the answers.
The second equation states essentially
that y = z. So let z = a (a
constant).
Substituting this into the first
equation
gives x + a - 3a = -1. So x - 2a = -1
and
x = 2a - 1, y
= a, and z = a.
Solution: (2a - 1, a, a) where a is any real number.
5. The same sort of thing is done when a system has fewer
equations
than variables.
Example:
2x - 3y + z = -2
-4x + 9y
= 7
Add 2 times the first
equation
to the second equation. 3y + 2z = -2. So 3y
= -2z - 2
and y = (-2/3) z - 2/3.
Put this value of y back into the second
equation:
- 4x + 9((-2/3)z - 2/3) = 7.
- 4x - 6z - 6 = 7
- 4x = 6z + 13
x =
(-3/2)z
- 13/4
Let z = a.
Solution Set: ( (-3/2)a - 13/4,
(-2/3)a - 2/3, a) where a is any real number.
Problems (there are no problems here yet)