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4|x-7|-6≤14 4|x-7|≤20 |x-7| <= 5 x-7<= 5 or x-7<= -5 x<=12 or x<=2
how did you get 20?
4|x-7|-6≤14 add 6 to both sides 4|x-7|≤20
devide both sides by 4 |x-7| <= 5
why do you divide by 4?
cuz i see 4|x-7|≤20 and i want to isolate x eventually therfore im interested in |x-7| on the left side.
i have 4 * |x-7| on the left side...im only interested in |x-7| so i hav to devide the left side by 4...since we devide the left side by 4, we HAVE to devide the right side by 4
ok, i see. how did you get x-7≤ 5 or x-7≤ -5? how'd you get the -5?
by definition, if |x| = A x = A or x = -A
cuz when u take the absolute value of A, it is A and when u take the absolute value of -A, it is A This 'A' can be any real number
so theres two answers then? x≤12 and x≤2
yes. x≤12 OR x≤2
ok, if i was to graph the solution, how would i?
i drew a line at x=2, the red vertival line..and just shaded the area (that is left side in this case) x<=2
do the same for x<=12 draw a line at x=12, and shade the less than area, that is left side
ok, i see. thank you.
I think that is wrong^ :|
|x-7|<=5 -5<= (x-7) <= 5 2<= x <= 12 CORRECT that is different from the other person answer
also, if the answer was x<=12 or x<=2 then you would merge both the solutions to just x<=12, bec ause that is where they both hold, BUT REMEMBER THAT ANSWER IS INCORRECT , corrrect answer is above
EDIT: you would merge them to x<=2