Answer:
x = -35/6 or 15/2
Explanation:
Remembering that |x| = x (x ≥ 0) or -x (x ≤ 0), we find each absolute value must be considered two ways, each with its own applicable domain.
The equation resolves to four equations with different domains.
- 2/5x +5 = 4/5x +2 . . . . 2/5x +5 ≥ 0 and 4/5x +2 ≥ 0
- 2/5x +5 = -(4/5x +2) . . . . 2/5x +5 ≥ 0 and 4/5x +2 ≤ 0
- -(2/5x +5) = 4/5x +2 . . . . 2/5x +5 ≤ 0 and 4/5x +2 ≥ 0
- -(2/5x +5) = -(4/5x +2) . . . 2/5x +5 ≤ 0 and 4/5x +2 ≤ 0
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The domains can be found by simplifying those equations.
2/5x +5 ?? 0 ⇒ x + 25/2 ?? 0 x ?? -25/2
4/5x +2 ?? 0 ⇒ x + 5/2 ?? 0 ⇒ x ?? -5/2
where ?? represents ≥ or ≤.
Domain 1: x ≥ -25/2 and x ≥ -5/2 ⇒ x ≥ -5/2
Domain 2: x ≥ -25/2 and x ≤ -5/2 ⇒ -25/2 ≤ x ≤ -5/2
Domain 3: x ≤ -25/2 and x ≥ -5/2 ⇒ the empty set
Domain 4: x ≤ -25/2 and x ≤ -5/2 ⇒ x ≤ -25/2
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We can simplify the equations by multiplying by 5 and subtracting one side from both sides:
5(2/5x +5) -5(4/5x +2) = 0 . . . . equations 1 or 4; domains 1 or 4
2x +25 -4x -10 = 0
-2x +15 = 0
x = 15/2 . . . . . in domain 1
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5(2/5x +5) +5(4/5x +2) = 0 . . . . equation 2; domain 2
2x +25 +4x +10 = 0
6x +35 = 0
x = -35/6 . . . . . in domain 2
The solutions are x = -35/6 or 15/2.
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Additional comment
I find the above consideration of variations of the equation and the respective domains to be tedious. I prefer to let a graphing calculator show the solution to f(x) = 0, where f(x) is the difference between the left and right sides of the original absolute value equation.