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How do you solve #8abs(x+7)-3=5#?

Answer 1

Eva Abramson

#x=-8" or "x=-6#

#"the expression inside the absolute value bars can be"# #"positive or negative hence we have to consider both"#

#"isolate "|x+7|" by adding 3 to both sides"#

#8|x+7|cancel(-3)cancel(+3)=5+3#

#rArr8|x+7|=8#

#"divide both sides by 8"#

#cancel(8)/cancel(8)|x+7|=8/8#

#rArr|x+7|=1#

#"consider the "color(magenta)" positive value ""of "x+7#

#rArrx+7=1larr"subtract 7 from both sides"#

#rArrx=1-7=-6#

#"consider the "color(magenta)"negative value ""of "x+7#

#rArr-(x+7)=1larr"distribute"#

#rArr-x-7=1larr"add 7 to both sides"#

#rArr-x=1+7=8#

#rArrx=-8#

#color(blue)"As a check"#

These values are the solutions if you substitute them into the left side of the equation and see if they equal the right side.

#x=-6to8|-6+7|-3=8-3=5#

#x=-8to8|-8+7|-3=8|-1|-3=8-3=5#

#rArrx=-8" or "x=-6" are the solutions"#

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Answer 2

Eva Adamson

To solve the equation 8| x + 7 | - 3 = 5, you would first add 3 to both sides, then divide by 8, and finally solve for the absolute value expression by considering both positive and negative cases. The solutions are x = -(\frac{9}{8}) and x = (\frac{13}{8}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.