How do you solve #-42x - 12= - 21x - 9#?

Question

500 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-30T02:40:05

See a solution process below:

First, add #color(red)(42x)# and #color(blue)(9)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#color(red)(42x) - 42x - 12 + color(blue)(9) = color(red)(42x) - 21x - 9 + color(blue)(9)#

#0- 3 = (color(red)(42) - 21)x - 0#

#-3 = 21x#

Now, divide each side of the equation by #color(red)(21)# to solve for #x# while keeping the equation balanced:

#-3/color(red)(21) = (21x)/color(red)(21)#

#-1/7 = (color(red)(cancel(color(black)(21)))x)/cancel(color(red)(21))#

#-1/7 = x#

#x = -1/7#