How do you express #\frac { 5} { 3a } + 4b = \frac { 1- 2c} { d } # in terms of #c#?

1 Answer
smendyka
Aug 9, 2017

See a solution process below:

Explanation:

Step 1) Multiply each side of the equation by #color(red)(d)# to eliminate the fraction on the side of the equation with the #c# variable while keeping the equation balanced:

#color(red)(d)(5/(3a) + 4b) = color(red)(d)((1 - 2c)/d)#

#(color(red)(d) xx 5/(3a)) + (color(red)(d) xx 4b) = cancel(color(red)(d))((1 - 2c)/color(red)(cancel(color(black)(d))))#

#(5d)/(3a) + 4bd = 1 - 2c#

Step 2) Subtract #color(red)(1)# from each side of the equation to isolate the #c# term while keeping the equation balanced:

#(5d)/(3a) + 4bd - color(red)(1) = -color(red)(1) + 1 - 2c#

#(5d)/(3a) + 4bd - 1 = 0 - 2c#

#(5d)/(3a) + 4bd - 1 = -2c#

Step 3) Divide each side of the equation by #color(red)(-2)# to solve for #c# while keeping the equation balanced:

#((5d)/(3a) + 4bd - 1)/color(red)(-2) = (-2c)/color(red)(-2)#

#(5d)/(color(red)(-2) xx (3a)) + (4bd)/color(red)(-2) - 1/color(red)(-2) = (color(red)(cancel(color(black)(-2)))c)/cancel(color(red)(-2))#

#-(5d)/(6a) - 2bd + 1/2 = c#

#c = -(5d)/(6a) - 2bd + 1/2#

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