Answer: x = 3 (choice A)
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Work Shown
4|x-3| + 1 = 1
4|x-3| = 1-1
4|x-3| = 0
|x-3| = 0/4
|x-3| = 0
x-3 = 0
x = 3
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Check:
Plug in x = 3 and simplify.
4|x-3| + 1 = 1
4|3-3| + 1 = 1
4*|0| + 1 = 1
4*0 + 1 = 1
0 + 1 = 1
1 = 1
We arrive at a true statement, so the first equation is true when x = 3.
The solution x = 3 has been confirmed.
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Let's see what happens if we plugged in x = 4
4*|x-3| + 1 = 1
4*|4-3| + 1 = 1
4*|1| + 1 = 1
4*1 + 1 = 1
4 + 1 = 1
5 = 1
The last statement is false, so the original equation is false when x = 4.
We rule out x = 4 as a solution.
You should find that x = 6 leads to a false equation as well, so x = 6 isn't a solution either.
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Another way to check is to use graphing software like GeoGebra or Desmos.
Plot the equation y = 4*abs(x-3)+1 where the "abs" refers to "absolute value". Also plot the horizontal line y = 1.
The two intersect at exactly one point and it is (3,1). The x coordinate of this intersection is the final answer.