Download 3-2 Solving Linear Systems Algebraically

3-2
Solving Linear Systems
Algebraically
Objective: CA 2.0: Students solve
system of linear equations in two
variables algebraically.
Substitution Method
1. Solve one equation for one of its
variables
2. Substitute the expression from
step 1 into the other equation and
solve for the other variable
3. Substitute the value from step 2
into the revised equation from step
1 and solve
Example 1:
Solve the linear system of equation
using the substitution method.
3x  4 y  4
x  2y  2
Step 1:
Solve one equation for one of its
variables.
x  2y  2
x  2 y  2
Step 2:
Substitute the expression from
step 1 into the other equation and
solve.
3 2 y  2  4 y  4
6 y  6  4 y  4
2 y  10
2 y  6  4
y 5
Step 3:
Substitute the value from step 2
into the revised equation from
step 1 and solve.
x  2 y  2
x  2  5  2
x  8
The solution is (-8, 5)
Which equation should you choose in step 1?
In general, you should solve for a variable whose
coefficient is 1 or -1
The Linear Combination
Method
Step 1: Multiply one or both of the
equations by a constant to obtain
coefficients that differ only in sign
for one variable.
Step 2: Add the revised equations
from step 1. Combining like terms
will eliminate one variable. Solve
for the remaining variable.
Step 3: Substitute the value
obtained in Step 2 into either of
the original equations and solve
for the other variable.
Example 2
Solve the linear system using
the Linear Combination
(Elimination) Method.
2 x  4 y  13
4x  5 y  8
Step 1:
Multiply one or both of the
equations by a constant to obtain
coefficients that differ only in sign
for one variable
2 x  4 y  13
4x  5 y  8
Multiply everything by -2
Leave alone
After step 1 we now have…
-4x + 8y = -26
4x – 5y = 8
Step 2:
Add the revised equations.
4 x  8 y  26
4 x  5 y  8
3 y  18
y  6
Solve for y
Step 3
Substitute the value obtained in
Step 2 into either of the original
equations and solve for the other
variable.
2 x  4  6  13
2x  24  13
2x  13  24
2x  11
11
x
2
Check your solution   11 , 6 



2

2 x  4 y  13
 11 
2     4  6   13
 2
11  24  13
The solution checks
Example 3:
Linear Combination: Multiply both Equations
Solve the linear system using the Linear
Combination method.
7 x  12 y  22
5 x  8 y  14
Step 1) Multiply one or both equations by a
constant to obtain coefficients that differ only in
sign for one variable.
2  7 x  12 y   2  22   14 x  24 y  44
3  5 x  8 y   3 14   15 x  24 y  42
  x  2
x2
Step 2) Add the revised equations
14 x  24 y  44
15 x  24 y  42
 x  2
x2
Step 3) Substitute the value obtained in Step 2
into either original equation
5  2   8 y  14
10  8 y  14
8 y  24
y3
Solution (2, 3)
Example 4
Linear Systems with many or no
solutions
Solve the linear system.
x  2y  3
2x  4 y  7
Use the substitution method
Step 1)
Step 2)
x  2y  3
x  2y  3
2  2 y  3  4 y  7
4y  6  4y  7
6=7
Because 6 is not equal to 7, there are no
solutions.
Two lines that do not intersect are parallel.
x  2y  3
2 y   x  3
x 3
y

2 2
x 3
y 
2 2
2x  4 y  7
4 y   2 x  7
2 x 7
y

4 4
x 7
y 
2 4
Solve the Linear System
6 x  10 y  12
15 x  25 y  30
Solve using the linear combination method
Step 1)
5  6 x  10 y   5 12 
2  15 x  25 y   2  30 
Step 2) Add revised equations
30 x  50 y  60
30 x  50 y  60
00  0
Because the equation 0 = 0 is always true, there
are infinitely many solutions.
Linear Equations that have infinitely many
solutions are equivalent equations for the same
line.
6 x  10 y  12
15 x  5 y  30
2  3 x  5 y   2  6  5  3 x  5 y   5  6 
3x  5 y  6
3x  5 y  6
Home work page 153 12 – 20 even, 24 – 34
even, 38 – 52 even.