Workbench∣x+1∣ is the distance from x to −1, and ∣x−3∣ is the distance from x to 3. Drag x so that the two distances add up to exactly 6. How many positions can you find?
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What this move does
Some equations do not look like ax=b at first, but become it after some rearranging. The most common ones contain absolute values. Remember one sentence:
∣x−p∣is the distance on the number line fromxtop.
To remove the absolute value signs, split the number line into pieces according to whether each inside is positive or negative:
Find the points where each expression inside an absolute value equals 0. They cut the number line into pieces.
In each piece, remove the absolute value signs to get an ordinary linear equation, and solve it.
Check: does the number you found lie in that piece? If not, reject it.
Predict first
How many solutions does ∣x−2∣=5 have?
Choose an answer first; then you can see the result.
✓You predicted it!✗Not what you expected — which is exactly why it is worth a pause.
There are two points at distance 5 from 2: one to the right (7) and one to the left (−3).
Drag the point around, then set the right-hand side c to 0 and to −1. How does the number of solutions change?
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Work through one
Example: solve |x + 1| + |x − 3| = 6
Step 1 / 7
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Find the cut points
x+1=0 gives x=−1, and x−3=0 gives x=3. These two points cut the number line into three pieces: left, middle and right.
x<−1,−1≤x≤3,x>3
The middle piece
The left piece
The right piece
All the solutions
Change 6 to 4
Now change it to 3
An application: repeating decimals as fractions
Which fraction is 0.272727⋯? Call it x. Move the decimal point two places to the right, and the repeating tail does not change at all:
100x=27.2727⋯=27+x
That is a linear equation. Solve it on the balance:
Turn 0.2727… into a fraction
Step 1 / 4
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Set up the equation
100 lots of x weigh the same as “27 plus one more x”.
100x=x+27
Subtract x from both sides
Divide both sides by 99
Check
When it fails
Break it
Counterexample lab: remove one condition
Conclusion: the equation ∣x−2∣=c has a solution← still true?
c=3: there are two points 3 units from 2. Click “Show all solutions” to check.
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Now c=−1. Drag wherever you like: can a distance ever be −1?
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A distance is never negative, so ∣x−2∣=−1 has no solution. Write x−2=±(−1) without looking at the sign of the right side and you “find” x=1 and x=3. Both are fake.
Practice
Exercise 1Basic
Solve ∣x+3∣=4.
Hint · direction
∣x+3∣ is the distance from x to which point?
Hint · method
x+3 equals 4 or −4.
Hint · first step
x+3=4 gives x=1. What about the other case?
Full solution
x+3=−4 gives x=−7. So x=1 or x=−7: both are at distance 4 from −3.
Exercise 2Basic
Write the repeating decimal 0.444⋯ as a fraction.
Hint · direction
Let x=0.444⋯. What should you multiply by so that the decimal part stays the same?
Hint · method
Only one digit repeats, so multiply by 10.
Hint · first step
10x=4.444⋯=4+x.
Full solution
9x=4, so x=94.
Exercise 3Advanced
Solve ∣x−1∣+∣x−5∣=10.
Hint · direction
When x is between 1 and 5, what is the sum of the two distances?
Hint · method
In the middle piece the sum is always 4, so there is no solution there. Remove the absolute value signs separately for x<1 and for x>5.
Hint · first step
When x>5: (x−1)+(x−5)=10, that is, 2x−6=10.
Full solution
When x>5, x=8; when x<1, (1−x)+(5−x)=10 gives x=−2. Both numbers lie in their own pieces, so x=−2 or x=8.
Exercise 4Challenge
Solve ∣3x−2∣=x+4.
Hint · direction
Split into two cases by the sign of 3x−2. What must you do after solving each case?
Hint · method
When 3x−2≥0 the equation is 3x−2=x+4; when 3x−2<0 it is 2−3x=x+4. After solving, check that each answer meets its own condition.
Hint · first step
First case: 2x=6, so x=3. Then 3x−2=7≥0, which fits.
Full solution
Second case: −4x=2, so x=−21, and then 3x−2=−27<0, which also fits. Check: when x=3 both sides are 7; when x=−21 both sides are 27. So x=3 or x=−21.
Writing only one case when removing an absolute value
✗
The wrong step: ∣x−2∣=5, so x−2=5 and x=7.
What went wrong? Think first, then open
DiagnosisTwo numbers have absolute value 5: 5 and −5. On the number line there are two points 5 units from 2, one on each side.
✓
The right step: x−2=5 or x−2=−5, so x=7 or x=−3.
“An absolute value is a distance, so look on both sides”
Exhibit
Taking every case's answer as a solution
✗
The wrong step: ∣2x−1∣=x−2. When 2x−1≥0: 2x−1=x−2, so x=−1. When 2x−1<0: 1−2x=x−2, so x=1. So x=−1 or x=1.
What went wrong? Think first, then open
DiagnosisEach result only counts inside its own piece. x=−1 does not satisfy 2x−1≥0, and x=1 does not satisfy 2x−1<0. Substituting back shows it too: when x=1 the left side is 1 and the right side is −1.
✓
The right step: Neither result lies in its own piece, so both are rejected: the original equation has no solution. (You could also see this at the start: the left side is at least 0, so we need x≥2, and then 2x−1=x−2 gives x=−1, a contradiction.)
“A number found in one piece must be checked against that piece”
Using and Rearranging Linear Equations ← → step · F full screen · Esc exit