solve for x: −3|2x + 6| = −12 a. x = −1 and x = −5 <--- correct answer b. no solution c. x = 1 and x = 5 d. x = −9 and x = 3

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solve for x: −3|2x + 6| = −12 a. x = −1 and x = −5 <--- correct answer b. no solution c. x = 1 and x = 5 d. x = −9 and x = 3

Algebra
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|dw:1434963289424:dw|
first you can divide both sides by -3 |dw:1434963323427:dw|
|dw:1434963334417:dw|

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|dw:1434963344378:dw|
make sense so far?
yes, wow thank you. i kept distributing
so you will have |2x+6| = 4
|2x+6| = 4 breaks down into these two equations 2x+6 = 4 or 2x+6 = -4
does the absolute value rule also apply here?
I thought.. it's 2x+6=4 or -2x-6= 4...or am I remembering this wrong?
either way works because -2x-6 = 4 is equivalent to 2x+6 = -4
oh ok! I prefer either we have 2x+6 =4 or -2x-6=4 whew
so now all we need to do is solve for x for those equations.
i got -5 twice???
oh there it is! For an absolute value equation, we split this equation into 2 situations. one where \[\abs 2x+6 \abs = 4\] Y U NO WORK LATEX NUGH! we have one situation where the 2x+6 is positive and another situation is the negative sign is distributed all over 2x+6
2x+ 6 = 4 -2x-6=4 should be our equations
let's start at 2x+6=4 solve for x
x = 5
x = -5 sorry
if x = -5 and I plug it back into the equation I have -4.. -4 doesn't equal to 4 for 2x+6=4
I need x by itself.. so what do we need to do?
subtract 6 on both sides, then divide by 2
yeah
\[2x+6=4 \] \[2x=4-6\] \[2x=-2\]
ohhh ok so x = -1
right so our next equation -2x-6=4
so what do we need to do to get x by itself ^^
@jillybean367 are you there? ^^
add 6 to both sides, then multiply by 2?
yes for addition part, but no for the multiplication \[-2x-6=4\] \[-2x-6+6=4+6\] \[-2x=10\]
divide by -2
mhm and what do we have?
x =5 ?
close you are dividing a positive number with a negative number example \[\frac{-12}{4} = -3\]
so in your problem we have \[x = \frac{10}{-2}\] or \[x = \frac{-10}{2}\]
what is x?
5?
the answer has a negative sign though...
we can't drop the negative sign.
oh. so its x = -5?
yes, so we have x = -1 and x = -5 if you plug those values back in the original equation you will have -12=-12 so those are the correct x values
Wow thank you so much!

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