How do you solve #40y + 40= 50y + 50#?

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Answer 1 · 2017-07-15T03:02:26

See a solution process below:

Step 1) Subtract #color(red)(40y)# and #color(blue)(50)# from each side of the equation to isolate the #y# term while keeping the equation balanced:

#-color(red)(40y) + 40y + 40 - color(blue)(50) = -color(red)(40y) + 50y + 50 - color(blue)(50)#

#0 - 10 = (-color(red)(40) + 50)y + 0#

#-10 = 10y#

Now, divide each side of the equation by #color(red)(10)# to solve for #y# while keeping the equation balanced:

#-10/color(red)(10) = (10y)/color(red)(10)#

#-1 = (color(red)(cancel(color(black)(10)))y)/cancel(color(red)(10))#

#-1 = y#

#y = -1#