Answer:
c = {-2/3, 1}
Explanation:
You want the solutions to the absolute value equation ...
|2c +8| = |10c|
Breakpoints
The equation's domain can be divided into parts that have boundaries where the argument of the absolute value function is zero. Here, those boundaries are ...
2c +8 = 0 ⇒ c = -8/2 = -4
10c = 0 ⇒ c = 0
Piecewise function
Based on these boundary values, we can rewrite the equation into three different equations in three different domains. The absolute value function negates a negative argument, so we can write ...
- -(2c +8) = -(10c) . . . . for c < -4
- 2c +8 = -(10c) . . . . for -4 ≤ c < 0
- 2c +8 = 10c . . . . for 0 ≤ c
c < -4
-(2c +8) = -(10c) . . . . equation definition in this domain
2c +8 = 10c . . . . . . . multiply by -1
8 = 8c . . . . . . . . . . . subtract 2c
1 = c . . . . . . . . . . . . divide by 8; no solution in the given domain
-4 ≤ c < 0
2c +8 = -10c . . . . equation definition in this domain
12c +8 = 0 . . . . . add 10c
c +2/3 = 0 . . . . . divide by 12
c = -2/3 . . . . . . . subtract 2/3
0 ≤ c
2c +8 = 10c . . . . equation definition in this domain
8 = 8c . . . . . . . . subtract 2c
1 = c . . . . . . . . . divide by 8
The solutions to the equation are c = -2/3 and c = 1.
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Additional comment
These equations are nicely solved by a graphing calculator. The equation can be rewritten to the form f(c) = 0 by subtracting one side from both sides. Then the solutions are the x=intercepts of the graph.
You will notice the breakpoints in the graph at x = -4 and x = 0, where the nature of the relation changes.