How do you solve #20c + 5= 10c + 25#?

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Answer 1 · 2017-07-15T12:18:08

See a solution process below:

Step 1) Subtract #color(red)(5)# and #color(blue)(10c)# from each side of the equation to isolate the #c# term while keeping the equation balanced:

#-color(blue)(10c) + 20c + 5 - color(red)(5) = -color(blue)(10c) + 10c + 25 - color(red)(5)#

#(-color(blue)(10) + 20)c + 0 = 0 + 20#

#10c = 20#

Step 2) Divide each side of the equation by #color(red)(10)# to solve for #c# while keeping the equation balanced:

#(10c)/color(red)(10) = 20/color(red)(10)#

#(color(red)(cancel(color(black)(10)))c)/cancel(color(red)(10)) = 2#

#c = 2#