Solving a Linear System of Linear
Equations in Three Variables by Substitution
The substitution method involves algebraic
substitution of one equation into a variable of the other.
This
will be the sample equation used through out the instructions:
Equation 1) x – 6y – 2z = -8
Equation 2) -x + 5y + 3z
= 2
Equation 3) 3x - 2y – 4z
= 18
Steps in order to solve systems of linear equations through
substitution:
- Solve
one of the equations for one of its variables. From the three variables,
there is no incorrect choice so choose to solve for any variable.
ü
i.e.: x= 6y +2z -8
- Next,
substitute the value from the first variable you solved for into the other
equation and solve for the next variable.
ü
i.e.: -(6y + 2z -8)
+2y-5z = -30
-y + z
+ 8 = 2
z = y –
6
- Substitute
the value from the two variables that you solved and plug it into the
remaining equation and solve for the last remaining variable. This step
should allow you to solve for a real number.
ü
i.e.: 3(6y + 2y – 8) – 2y – 4 (y – 6) = 18
18y – 36 = 18
18y = 54
y = 3
- After
solving for the final variable, plug in the value of the most recent
variable that you found (in terms of the example, y=3) into the answer of
another equations with variables remaining (in terms of the example, z = y
– 6, x = 6). Note: Preferably,
plug in the value to the most simplified equation.
ü
i.e.: z = 3 – 6
z = -3
- After
solving for another variable, you should have the remaining pieces of the
puzzle for the last equation.
ü
i.e.: x = 6(3) + 2(-3) – 8
x = 18 – 6 – 8
x = 4
- Therefore,
in the end, you will have successfully have found the answers to a system
of linear equations in three variables.
ü
Answer = (4, 3, -3)
PRACTICE PROBLEMS
Practice and
fine-tune your substitution skills!
Good Luck!
- -2x +
y + 6z = 1
3x + 2y + 5z = 16
7x + 3y – 4z = 11
- x – 3y
+ 6z = 21
3x + 2y -5z = -30
2x – 5y +
2z = -6
- x – 6y
– 2z = -8
-x + 5y + 3y = 2
3x – 2y – 4z = 18
Challenge Problem
Hint: Stay open-minded
3x + 3y +z = 30
10x - 3y - 7z = 17
-6y + 7y
+ 3z = -49