How do you solve #\frac{251}{6} = \frac{1}{2} - 4( \frac{8}{3} k - 1)#?

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Answer 1 · 2016-09-16T09:10:18

#k = - (7) / (2)#

We have: #(251) / (6) = (1) / (2) - 4 ((8) / (3) k - 1)#

First, let's subtract #(1) / (2)# from both sides of the equation:

#=> (251) / (6) - (1) / (2) = - 4 ((8) / (3) k - 1)#

#=> (496) / (12) = - 4 ((8) / (3) k - 1)#

Then, let's divide both sides by #- 4#:

#=> (496) / (- 48) = (8) / (3) k - 1#

#=> (8) / (3) k - 1 = - (496) / (48)#

#=> (8) / (3) k - 1 = - (31) / (3)#

Now, let's add #1# to both sides:

#=> (8) / (3) k = - (31) / (3) + 1#

#=> (8) / (3) k = - (28) / (3)#

Next, let's multiply both sides by #3#:

#=> 8 k = - 28#

Finally, to solve for #x#, let's divide both sides by #8#:

#=> k = - (28) / (8)#

#=> k = - (7) / (2)#

Answer 2 · 2016-09-16T09:10:58

#k = -4.25#

In an equation with fractions, we can get rid of the denominators by multiplying through by the LCM of the denominators.

In this case #LCM =color(red)(6)" "rarr#multiply each term by 6

#color(red)(6)xx 251/6 =color(red)(6)xx 1/2 - color(red)(6)xx4(8/3 k - 1)#

Cancel the denominators

#cancelcolor(red)(6)xx 251/cancel6 =cancelcolor(red)(6)^3xx 1/cancel2 - cancelcolor(red)(24)^8xx8/cancel3 k - color(red)(24)xx1#

#251 = 3- 64k -24#

#64k = 3-24-251#

#64k = -272" "larr# divide by 64

#k = -4.25#