|x-1|+|x+1|=1
1) We solve in :
The absolute-value arguments are zero at . Split ℝ at these points and solve on each interval
On : ; therefore, ; , so
On , the equation becomes
2) Let's solve in :
ⓘ Start from and move back on the number line, down to zero: = ■Group like terms by adding their coefficients: =
ⓘ Adding 0 changes nothing: =
ⓘ Divide both sides by
ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator: =
Solution found:
As , we cannot keep the solution
On : ; therefore, ; ; therefore,
On , the equation becomes
3) We solve in :
ⓘ A minus sign before a parenthesis changes the sign of each of its terms: =
ⓘ Start from and move forward on the number line: = ■Group like terms by adding their coefficients: =
ⓘ Adding 0 preserves the original value: =
The statement is false, whatever the value of
Solution:
On : ; therefore, ; ; therefore,
On , the equation becomes
4) We solve in :
ⓘ Start from and move forward on the number line, up to zero: = ■Like terms are grouped by adding their coefficients: =
ⓘ Adding 0 changes nothing: =
ⓘ Divide both sides by
Solution obtained:
Since , we cannot keep the solution
▶Solution:
