How do you solve #-4= \frac { 2} { 4} + b#?

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Answer 1 · 2017-12-05T03:22:00

See a solution process below:

First, multiply each side of the equation by #color(red)(2)# to eliminate the fraction while keeping the equation balanced:

#color(red)(2) xx -4 = color(red)(2)(2/4 + b)#

#-8 = (color(red)(2) xx 2/4) + (color(red)(2) xx b)#

#-8 = 4/4 + 2b#

#-8 = 1 + 2b#

Next, subtract #color(red)(1)# from each side of the equation to isolate the #b# term while keeping the equation balanced:

#-8 - color(red)(1) = 1 - color(red)(1) + 2b#

#-9 = 0 + 2b#

#-9 = 2b#

Now, divide each side of the equation by #color(red)(2)# to solve for #b# while keeping the equation balanced:

#-9/color(red)(2) = (2b)/color(red)(2)#

#-9/2 = (color(red)(cancel(color(black)(2)))b)/cancel(color(red)(2))#

#-9/2 = b#

#b = -9/2#