Question 134463
A) 



{{{2*abs(3-p)-4=2}}} Start with the given equation



{{{2*abs(3-p)=6}}} Add 4 to both sides.



{{{abs(3-p)=3}}} Divide both sides by 2




Break up the absolute value (remember, if you have {{{abs(x)=a}}}, then {{{x=-a}}} or {{{x=a}}})


{{{3-p=-3}}} or {{{3-p=3}}} Set the expression {{{3-p}}} equal to the original value 3 and it's opposite -3





Now lets focus on the first equation  {{{3-p=-3}}}



{{{-p=-3-3}}}Subtract 3 from both sides



{{{-p=-6}}} Combine like terms on the right side



{{{p=(-6)/(-1)}}} Divide both sides by -1 to isolate p




{{{p=6}}} Divide





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Now lets focus on the second equation {{{3-p=3}}}




{{{-p=3-3}}}Subtract 3 from both sides



{{{-p=0}}} Combine like terms on the right side



{{{p=(0)/(-1)}}} Divide both sides by -1 to isolate p




{{{p=0}}} Divide






So the solutions to {{{2*abs(3-p)-4=2}}} are:


{{{p=6}}} or {{{p=0}}}




Notice if we graph  {{{y=2*abs(3-x)-4}}} and {{{y=2}}} (just set each side equal to y and graph), we get



{{{graph(500,500,-2,8,-10,10,2*abs(3-x)-4,2)}}}  Graph of {{{y=2*abs(3-x)-4}}} (red) and {{{y=2}}}(green)


and we can see the two graphs intersect at {{{x=6}}} and {{{x=0}}}. So this verifies our answer.





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{{{2*abs(3-p)-4<2}}} Start with the given inequality



{{{2*abs(3-p)<6}}} Add 4 to both sides.



{{{abs(3-p)<3}}} Divide both sides by 2




Break up the absolute value (remember, if you have {{{abs(x)< a}}}, then {{{x > -a}}} and {{{x < a}}})


{{{3-p > -3}}} and {{{3-p < 3}}} Break up the absolute value inequality using the given rule



{{{-3 < 3-p < 3}}} Combine the two inequalities to get a compound inequality



{{{-6 < -p < 0}}} Subtract 3 from all sides



{{{6 > p > 0}}} Multiply all sides by -1 to isolate p. This will flip the inequality signs


{{{0<p<6}}} Rearrange the inequality


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Answer:


So our answer is


{{{0<p<6}}}




which looks like this in interval notation



*[Tex \LARGE \left(0,6\right)]



if you wanted to graph the solution set, you would get

{{{drawing(500,50,-10,10,-10,10,
number_line( 500, -7, 13),

blue(line(-2.5,-7,2.65,-7)),
blue(line(-2.5,-6,2.65,-6)),
blue(line(-2.5,-5,2.65,-5)),

circle(-3,-5.8,0.35),
circle(-3,-5.8,0.4),
circle(-3,-5.8,0.45),


circle(3,-5.8,0.35),
circle(3,-5.8,0.4),
circle(3,-5.8,0.45)




)}}} Graph of the solution set in blue and the excluded values represented by open circles