Elimination Strategies

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Example 1

{ 3 a + 5 b = 31 3 a + b = 23

step 1: Decide to cancel either a or b

For a: The coefficients 3 and 3 cancel perfectly through subtraction

For b: The coefficients 5 and 1 do not cancel easily.

We will choose to cancel a using subtraction

{ 3 a + 5 b = 31 3 a + ( 1 ) b = 23

step 2:
Perform addition/subtracting

We will subtract because we want to cancel 3a with 3a

3 a + 5 b = 31 3 a + ( 1 ) b = 23 0 a + 4 b = 8

step 3:
Using the equation from step 2, we will isolate b

0 a + 4 b = 8 4 b = 8 4 b 4 = 8 4 b = 2

step 4:
Using the value of a from step 3, we can solve for b using either of the original equations

We do this by substituting a with its value.

opition: a

3 a + 5 b = 31 3 a + 5 ( 2 ) = 31 3 a + 10 = 31 3 a = 21 a = 7

option b:

3 a + b = 23 3 a + ( 2 ) = 23 3 a = 21 a = 7


Final Answer:

{ a = 7 b = 2

Example 2

{ 5 x + 3 y = 5 5 x 5 y = 7

step 1: Decide how to cancel either x or y

For x: the coefficients of 5 and -5 can cancel perfectly through addition

For y: the coefficients of 3 and -5 do not cancel easily

we will choose to cancel the x terms through addition

{ 5 x + 3 y = 5 5 x 5 y = 7

step 2:
Perform addition/subtraction

We will do an addition because we want to cancel -5x with 5x

5 x + 3 y = 5 + ( 5 x 5 y = 7 ) 0 x 2 y = 12

step 3:
Using the equation from step 2, we will isolate y

0 x 2 y = 12 2 y = 12 2 y 2 = 12 2 y = 6

Step 4:

Using the value of y we found in step 3, we can solve for x using either of the original equations

We do this by replacing y with its value

option a:

5 x + 3 y = 5 5 x + 3 ( 6 ) = 5 5 x + 18 = 5 5 x = 23 x = 23 5  or  23 5

option b

5 x 5 y = 7 5 x 5 ( 6 ) = 7 5 x + 30 = 7 5 x = 23 x = 23 5 23 5

Final Answer:

{ x = 23 5 y = 6

Example 3

{ 2 f + 4 g = 2 3 f 5 g = 41

step 1: Adjusting the equations

For f: the coefficients of 2 and 3 do not easily cancel

For g: the coefficients of 4 and -5 do not cancel easily through addition or subtraction.

We can choose to adjust the equations so we can adjust either the f terms or the g terms. In this example, we will cancel the f terms.

We multiply both sides of the first equation by 3

3 ( 2 f + 4 g ) = ( z ) 3 6 f + 12 g = 6

Now we will multiply both sides of the second equation by 2

2 ( 3 f 5 g ) = ( 41 ) 2 6 f 10 g = 82

step 2:
Now our adjusted equations have coefficients for the f terms of 6f and 6f which can be cancelled through subtraction

6 f + 12 g = 6 6 f 10 g = 82 0 f + 22 g = 88

step 3:
From here we can solve for g

0 f + 22 g = 88 22 g = 88 22 g 22 = 88 22 g = 4

step 4: Using the value of g from step 3 we can solve for f

option a:

2 f + 4 g = 2 2 f + 4 ( 4 ) = 2 2 f + 16 = 2 2 f = 14 f = 7

option b:

3 f 5 g = 41 3 f 5 ( 4 ) = 41 3 f 20 = 41 3 f = 21 f = 7

Final Answer:

{ f = 7 g = 4