How do you solve #-(-3-q) = -1/2(6q - 26)#?

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Answer 1 · 2017-09-19T02:40:15

See a solution process below:

First, expand the terms in parenthesis on both sides of the equation being careful to handle the signs correctly:

#3 + q = (-1/2 xx 6q) - (-1/2 xx 26)#

#3 + q = (-6/2q) - (-26/2)#

#3 + q = -3q - (-13)#

#3 + q = -3q + 13#

Next, subtract #color(red)(3)# and add #color(blue)(3q)# to each side of the equation to isolate the #q# term while keeping the equation balanced:

#-color(red)(3) + 3 + q + color(blue)(3q) = color(blue)(3q) - 3q + 13 - color(red)(3)#

#0 + 1q + color(blue)(3q) = 0 + 10#

#(1 + color(blue)(3))q = 10#

#4q = 10#

Now, divide each side of the equation by #color(red)(4)# to solve for #q# while keeping the equation balanced:

#(4q)/color(red)(4) = 10/color(red)(4)#

#(color(red)(cancel(color(black)(4)))q)/cancel(color(red)(4)) = (2 xx 5)/color(red)(2 xx 2)#

#q = (color(red)(cancel(color(black)(2))) xx 5)/color(red)(color(black)(cancel(color(red)(2))) xx 2)#

#q = 5/2#