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Transcript
How to solve linear absolute value equations
Solving an equation with an absolute value is a question that comes up in many classes. This
worksheet will show you how to solve these equations.
Example 1 Solve the equation |π‘₯ βˆ’ 3| = 4
STEP 1: Clear the absolute value of anything that is attached to it ( i.e. a number
you that is added or subtracted to the absolute value, a number that is in front of
the absolute value). One side should have the absolute value expression and the
other side should be a number.
The equation |π‘₯ βˆ’ 3| = 4 does not have any numbers added or subtracted to the absolute
value, or any numbers in front of the absolute value. We will skip this step.
STEP 2: Once you have taken care of step 2, look at the number on the opposite
side of the absolute value and follow the guideline below.
If it is positive, go to STEP 3. If not, look at the following guideline.
If it’s negative, STOP !! . You are done with the problem. There is no solution( an
absolute value cannot equal a negative number!!).
If the number is 0, then take what is inside that absolute value and set it equal to
zero. You will only have one solution in this case.
With the equation |π‘₯ βˆ’ 3| = 4, because 4 is positive, we can go to Step 3.
STEP 3: You will now make two equations using the property that if
|π‘Œ| = number, then this it is equivalent to π‘Œ= number or π‘Œ= - number.
With |π‘₯ βˆ’ 3| = 4, we will make two equations:
Created by Ivan Canales, Senior Tutor. San Diego City College. [email protected]
π‘₯ βˆ’ 3 = 4 and π‘₯ βˆ’ 3 = βˆ’4
STEP 4: Solve the equations
π‘₯βˆ’3=4
π‘₯ βˆ’ 3 = βˆ’4
π‘₯=7
π‘₯ = βˆ’1
π‘₯βˆ’3+3=4+3
π‘₯ βˆ’ 3 + 3 = βˆ’4 + 3
Our solutions will be π‘₯ = 7, βˆ’1
Example 2 Solve the equation βˆ’4|3π‘₯ βˆ’ 3| + 2 = βˆ’4
STEP 1 (as above):
I will subtract 2 from both sides and divide by -4 to get the absolute value by itself.
βˆ’4|3π‘₯ βˆ’ 3| + 2 βˆ’ 2 = βˆ’4 βˆ’ 2
βˆ’4|3π‘₯βˆ’3|
βˆ’4
=
|3π‘₯ βˆ’ 3| =
STEP 2 (as above):
3
βˆ’6
βˆ’4
3
2
3
With the equation |3π‘₯ βˆ’ 3| = 2 , because 2 is positive, we can go to Step 3.
STEP 3 (as above):
3
3
With |3π‘₯ βˆ’ 3| = 2 we will make two equations: 3π‘₯ βˆ’ 3 = 2 and 3π‘₯ βˆ’ 3 =
STEP 4 (as above):
3
3π‘₯ βˆ’ 3 = 2
9
3π‘₯ = 2
3π‘₯ βˆ’ 3 =
3π‘₯ =
3
βˆ’3
2
2
Created by Ivan Canales, Senior Tutor. San Diego City College. [email protected]
βˆ’3
2
3
π‘₯=2
Our solutions will be π‘₯ =
3
2
1
,
π‘₯=2
1
2
Example 3 Solve the equation 3|5π‘₯ + 1| + 5 = 2
STEP 1 (as above):
I will subtract 5 from both sides and divide by 3 to get the absolute value by itself.
3|5π‘₯ + 1| + 5 βˆ’ 5 = 2 βˆ’ 5
3|5π‘₯ + 1| βˆ’3
=
3
3
|5π‘₯ + 1| = βˆ’1
STEP 2 (as above):
Because the number on the opposite side of the absolute value is βˆ’1, we will stop. There is
no solution.
Example 3 Solve the equation |2π‘₯ + 3| + 2 = 2
STEP 1 (as above):
I will subtract 2 from both sides.
|2π‘₯ + 3| + 2 βˆ’ 2 = 2 βˆ’ 2
STEP 2 (as above):
|2π‘₯ + 3| = 0
Since the number on the opposite side of the absolute value is 0, we the expression inside the
absolute value equal to zero.
2π‘₯ + 3 = 0
2π‘₯ + 3 βˆ’ 3 = 0 βˆ’ 3
Created by Ivan Canales, Senior Tutor. San Diego City College. [email protected]
2π‘₯ = βˆ’3
Our only solution will be π‘₯ =
βˆ’3
π‘₯=
βˆ’3
2
2
If you have any further questions about anything you read in this guide, please
ask a tutor for further assistance.
Created by Ivan Canales, Senior Tutor. San Diego City College. [email protected]