Elimination Strategies

Example 1

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

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:

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

${ a = 7 b = 2$

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

$0 x − 2 y = 12 − 2 y = 12 − 2 y − 2 = 12 − 2 y = − 6$

option b

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

${ x = − 23 5 y = − 6$

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:

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

${ f = − 7 g = 4$