How do you solve #- 2r + 6( r - 1) + 3r - ( 4- r ) = - ( r + 5)#?

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Answer 1 · 2017-01-28T19:45:50

See the entire solution process below:

First, expand the terms within parenthesis on both sides of the equation:

#-2r + 6r - 6 + 3r - 4 + r = -r - 5#

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

#-2r + 6r + 3r + r - 6 - 4 = -r - 5#

#(-2 + 6 + 3 + 1)r - 10 = -r - 5#

#8r - 10 = -r - 5#

Then, add #color(red)(r)# and #color(blue)(10)# to each side of the equation to isolate the #r# terms while keeping the equation balanced:

#8r - 10 + color(red)(r) + color(blue)(10) = -r - 5 + color(red)(r) + color(blue)(10)#

#8r + color(red)(r) - 10 + color(blue)(10) = -r + color(red)(r) - 5 + color(blue)(10)#

#9r - 0 = 0 + 5#

#9r = 5#

Now, divide each side of the equation by #color(red)(9)# to solve for #r# while keeping the equation balanced:

#(9r)/color(red)(9) = 5/color(red)(9)#

#(color(red)cancel(color(black)(9))r)/cancel(color(red)(9)) = 5/9#

#r = 5/9#