abs(2*x-1)+3=abs(5-x)+4
1) Let's solve in :
ⓘ Start from and move back on the number line, going below zero: =
The absolute-value arguments are zero at . Split ℝ at these points and solve on each interval
On : , so ; , so
On , the equation becomes
2) Let's solve in :
ⓘ Start from and move back on the number line, down to zero: = ■Like terms are grouped by adding their coefficients: =
ⓘ Adding 0 changes nothing: =
ⓘ We add to both sides
Solution obtained:
As , we can keep the solution
On : , so ; , so
On , the equation becomes
3) Let's solve in :
ⓘ Start from and move back more on the number line, away from zero: = ■A minus sign before a parenthesis changes the sign of each of its terms: =
ⓘ Start from and move back more on the number line, away from zero: = ■Like terms are grouped by adding their coefficients: =
ⓘ We add to both sides
ⓘ Both sides are divided by
Solution found:
As , we can keep the solution
On : , so ; , so
On , the equation becomes
4) Let's solve in :
ⓘ Start from and move back more on the number line, away from zero: = ■Like terms are grouped by adding their coefficients: =
ⓘ Start from and move forward on the number line, going past zero: =
ⓘ We subtract from both sides
Solution obtained:
As , we cannot keep the solution
▶Solutions obtained:
5) The numeric value of: is:
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