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Lesson 2.2
Solving Systems of Three Equations
Algebraically
1. When we have 3 equations in a system, we
can use the same 2 methods to solve them
algebraically as with 2 equations.
2. Whether you use substitution or elimination,
you should begin by numbering the
equations!
Solving Systems of Three Equations
3. Substitution Method (watch a clip)
a) Choose one of the 3 equations, isolate one
of the variables.
b) Substitute the “new” expression into each
of the other 2 equations.
c) These 2 equations now have only 2 variables
each (the same 2 variables). Solve this 2 x 2
system as before.
d) Find the third variable by substituting the 2
known values into any equation.
Solving Systems of Three Equations
4. Linear Combination Method
a) Choose 2 of the equations and eliminate one
variable as before.
b) Now choose one of the equations from step 1
and the other equation you didn’t use and
eliminate the same variable.
c) You should now have 2 equations (one from
step 1 and one from step 2) that you can solve
by elimination.
d) Find the third variable by substituting the 2
known values into any equation.
Solve the system of equations by substitution.
2x = -6y
x + y + z = 10
-4x – 4y – z = -4
You can easily solve the first equation for x.
2x = -6y
x = -3y
Then substitute –3y for x in each of the other two equations. Simplify each
equation.
x + y + z = 10
-3y + y + z = 10
x = -3y
-2y + z = 10
-4x – 4y – z = -4
-4(-3y) – 4y – z = -4
8y – z = -4
x = -3y
Solve –2y + z = 10
-2y + z= 10
z = 10 + 2y
for z.
Substitute 10 + 2y for z in 8y – z = -4. Simplify.
8y – z = -4
8y – (10 + 2y) = -4
6y – 10 = -4
y=1
z = 10 + 2y
Now, find the values of z and x.
Use z = 10 + 2y and x = -3y. Replace y with 1.
z = 10 + 2y
z = 10 + 2(1) y = 1
z = 12
x = -3y
x = -3(1)
x = -3
y=1
The solution is x = -3, y = 1, and z = 12.
Check each value in the original system.
Solve the system of equations by elimination.
(1) x + 4y – z = 20
(2) 3x + 2y + z = 8
(3) 2x – 3y + 2z = -16
To eliminate x using the first and
second equations, multiply each
side of the first equation by –3.
-3(x + 4y – z) = -3(20)
-3x – 12y + 3z = -60
To eliminate x using the first and
third equations, multiply each side
of the first equation by –2.
-2(x + 4y – z) = -2(20)
-2x – 8y + 2z = -40
Then add that result to the second
equation.
-3x–12y+ 3z = -60
3x +2y+ z = 8
-10y +4z = -52
Then add that result to the third
equation.
-2x – 8y + 2z = -40
2x – 3y +2z = -16
-11y + 4z = -56
Now you have two linear equations in two variables.
Solve this system.
Eliminate z by multiplying each side
of the second equation by –1 and adding the two equations.
 10 y  4 z  52
11y  4z  56
______________
y4
By substituting the value of y into one of the equations in two
variables, we can solve for the value of z.
-10y + 4z = -52
-10(4) + 4z = -52
y=4
4z = -12
z = -3 The value of z is –3.
Finally, use one of the original equations to find the
value of x.
x + 4y – z = 20
x + 4(4) – (-3) = 20
y = 4, z = -3, x= 1
The solution is x = 1, y = 4, and z = -3.
This can be written as the ordered triple (1, 4, -3).