3-23-2Solving Linear Systems Solving Linear Systems

AlgebraicallyAlgebraically

Objective: CA 2.0: Students solve system of linear equations in two

variables algebraically.

Substitution MethodSubstitution 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:Example 1:

Solve the linear system of equation using the substitution method.

3 4 4

2 2

x y

x y

Step 1:Step 1:

Solve one equation for one of its variables.

2 2x y

2 2x y

Step 2: Step 2:

Substitute the expression from step 1 into the other equation and solve.

3 2 2 4 4y y

6 6 4 4y y 2 6 4y

2 10y

5y

Step 3: Step 3: Substitute the value from step 2 into the revised equation from step 1 and solve.

2 2x y

2 5 2x

8x 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 The Linear Combination MethodMethod

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 2Example 2

Solve the linear system using the Linear Combination (Elimination) Method.

2 4 13

4 5 8

x y

x y

2 4 13

4 5 8

x y

x y

Step 1:Step 1:

Multiply one or both of the equations by a constant to obtain coefficients that differ only in sign for one variable

2 4 13

4 5 8

x y

x y

Multiply everything by -2

Leave alone

Step 2:Step 2:

Add the revised equations.

4 8 26

4 5 8

3 18

6

x y

x y

y

y

After step 1 we now have…

-4x + 8y = -264x – 5y = 8

Solve for y

Step 3Step 3 Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable.

2 4 6 13x

2 24 13x 2 13 24x

2 11x 11

2x

Check your solution 11, 6

2

2 4 13

112 4 6 13

2

11 24 13

x y

The solution checks

Example 3:

Linear Combination: Multiply both Equations

Solve the linear system using the Linear Combination method.

7 12 22x y 5 8 14x y

Step 1) Multiply one or both equations by a constant to obtain coefficients that differ only in sign for one variable.

2 7 12 2 22 14 24 44

3 5 8 3 14 15 24 42

x y x y

x y x y

2x 2x

Step 2) Add the revised equations

14 24 44x y 15 24 42x y

2

2

x

x

Step 3) Substitute the value obtained in Step 2 into either original equation

5 2 8 14

10 8 14

8 24

3

y

y

y

y

Solution (2, 3)

Example 4

Linear Systems with many or no solutions

Solve the linear system.

2 3

2 4 7

x y

x y

Use the substitution method

Step 1) 2 3

2 3

x y

x y

Step 2) 2 2 3 4 7y y 4 6 4 7y y

6 = 7

Because 6 is not equal to 7, there are no solutions.

Two lines that do not intersect are parallel.

2 4 7

4 2 7

2 7

4 47

2 4

x y

y x

xy

xy

2 3

2 3

3

2 23

2 2

x y

y x

xy

xy

Solve the Linear System 6 10 12

15 25 30

x y

x y

Solve using the linear combination method

Step 1)

5 6 10 5 12

2 15 25 2 30

x y

x y

Step 2) Add revised equations

30 50 60

30 50 60

x y

x y

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 10 12

2 3 5 2 6

3 5 6

x y

x y

x y

15 5 30

5 3 5 5 6

3 5 6

x y

x y

x y

Home work page 153 12 – 20 even, 24 – 34 even, 38 – 52 even.