How do you solve #-27= \frac { 3v - 6} { 8} - \frac { 8v + 9} { 2}#?

3 Answers
Aug 1, 2018

See a solution process below:

Explanation:

First, multiply each side of the equation by the appropriate form of #1# to eliminate the fractions while keeping the equation balanced:

#color(red)(16) xx -27 = color(red)(16)((3v - 6)/8 - (8v + 9)/2)#

#-432 = (color(red)(16) xx (3v - 6)/8) - (color(red)(16) xx (8v + 9)/2)#

#-432 = (cancel(color(red)(16))color(red)(2) xx (3v - 6)/color(red)(cancel(color(black)(8)))) - (cancel(color(red)(16))color(red)(8) xx (8v + 9)/color(red)(cancel(color(black)(2))))#

#-432 = color(red)(2)(3v - 6) - color(red)(8)(8v + 9)#

#-432 = (color(red)(2) xx 3v) - (color(red)(2) xx 6) - (color(red)(8) xx 8v) + (color(red)(8) xx 9)#

#-432 = 6v - 12 - 64v - 72#

Next, group and combine like terms on the right side of the equation:

#-432 = 6v - 64v - 12 - 72#

#-432 = (6 - 64)v + (-12 - 72)#

#-432 = -58v + (-84)#

#-432 = -58v - 84#

Then, add #color(red)(84)# to each side of the equation to isolate the #v# term while keeping the equation balanced:

#-432 + color(red)(84) = -58v - 84 + color(red)(84)#

#-348 = -58v - 0#

#-348 = -58v#

Now, divide each side of the equation by #color(red)(-58)# to solve for #v# while keeping the equation balanced:

#(-348)/color(red)(-58) = (-58v)/color(red)(-58)#

#6 = (color(red)(cancel(color(black)(-58)))v)/cancel(color(red)(-58))#

#6 = v#

#v = 6#

Aug 1, 2018

#v=6#

Explanation:

#(3v-6)/8-(8v+9)/2=-27#

#(3v-6-32v-36)/8=-27#

#(-29v-42)/8=-27#

#-29v-42=-216#

#-29v=-174#

#v=(-174)/(-29)=6#

Aug 1, 2018

Multiply throughout by 8

#-216=(3v-6) - 4(8v+9)#

Remove the brackets

#-216=3v-6-32v-36#

Collect like terms

#-216=-29v-42#

Add #29v and 216# to both sides

#29v=174#

Divide by 29

#v=174/29#

#v=6#