Question 958406
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The other tutors solve the problems by considering different cases; that is the standard formal algebraic method.<br>
For problems (a), (c), and (d) -- in which there is an absolute value expression on only one side -- the solutions can be found by a different method.<br>
The statement {{{abs(x-a)=b}}} can be interpreted to say "the difference between (variable) x and (fixed) a is equal to b".<br>
For (c), this can lead directly to the answer.  The equation {{{abs(x-4)=3}}} says that the difference between x and 4 is equal to 3, which means the values of x are 4-3=1 and 4+3=7.  Then since the statement is in fact a strict "less than" inequality, 1 and 7 are the boundaries of the solution set, so the solution in interval notation is (1,7).<br>
For (a) and (d) we need to multiply the whole inequalities by appropriate constants to make "x" appear by itself in the absolute value expression.<br>
(a)
{{{abs(2x-7)=3}}}
{{{abs(x-3.5)=1.5}}}<br>
The solutions are the two numbers for which the difference between the number and 3.5 is 1.5 -- i.e., 3.5-1.5 = 2 and 3.5+1.5 = 5.<br>
(d)
{{{abs(x/3-1)>=2}}}
{{{abs(x-3)>=6}}}<br>
The boundaries of the solution set are the two numbers whose difference from 3 is 6 -- i.e., 3-6 = -3 and 3+6 = 9.  Then since the statement is an inclusive "greater than" inequality, the solution set in interval notation is (-infinity,-3] U [9,infinity).<br>