How do you solve #4r - 5= 1- 2r#?

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Answer 1 · 2017-07-20T01:41:56

See a solution process below:

Step 1) Add #color(red)(5)# and #color(blue)(2r)# to each side of the equation to isolate the #r# term while keeping the equation balanced:

#4r - 5 + color(red)(5) + color(blue)(2r) = 1 - 2r + color(red)(5) + color(blue)(2r)#

#4r + color(blue)(2r) - 5 + color(red)(5) = 1 + color(red)(5) - 2r + color(blue)(2r)#

#(4 + color(blue)(2))r - 0 = 6 - 0#

#6r = 6#

Step 2) Divide each side of the equation by #color(red)(6)# to solve for #r# while keeping the equation balanced:

#(6r)/color(red)(6) = 6/color(red)(6)#

#(color(red)(cancel(color(black)(6)))r)/cancel(color(red)(6)) = 1#

#r = 1#