How do you solve #-x + 2= 6x - 3#?

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Answer 1 · 2017-06-08T02:10:20

See a solution process below:

Add #color(red)(x)# and #color(blue)(3)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#color(red)(x) - x + 2 + color(blue)(3) = color(red)(x) + 6x - 3 + color(blue)(3)#

#0 + 5 = color(red)(1x) + 6x - 0#

#5 = (color(red)(1) + 6)x#

#5 = 7x#

Now, divide each side of the equation by #color(red)(7)# to solve for #x# while keeping the equation balanced:

#5/color(red)(7) = (7x)/color(red)(7)#

#5/7 = (color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7))#

#5/7 = x#

#x = 5/7#