How do you solve #9= - 3( r - 6) - ( 3- 2r )#?

1 Answer
smendyka
Apr 9, 2017

See the entire solution process below:

Explanation:

First, expand the term in parenthesis on the left of the right side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#9 = color(red)(-3)(r - 6) - (3 - 2r)#

#9 = (color(red)(-3) xx r) - (color(red)(-3) xx 6) - (3 - 2r)#

#9 = -3r - (-18) - (3 - 2r)#

#9 = -3r + 18 - (3 - 2r)#

Next, remove the other term on the right side from parenthesis being careful to manage the signs of the individual terms correctly. Then group and combine like terms on the right side of the equation:

#9 = -3r + 18 - 3 + 2r#

#9 = -3r + 2r + 18 - 3#

#9 = (-3 + 2)r + (18 - 3)#

#9 = -1r + 15#

#9 = -r + 15#

Then subtract #color(red)(15)# from each side of the equation to isolate the #r# term while keeping the equation balanced:

#9 - color(red)(15) = -r + 15 - color(red)(15)#

#-6 = -r + 0#

#-6 = -r#

Now, multiply each side of the equation by #color(red)(-1)# to solve for #r# while keeping the equation balanced:

#color(red)(-1) xx -6 = color(red)(-1) xx -r#

#6 = r#

#r = 6#

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