How do you solve #-4/5x+6=x-3#?

1 Answer
May 23, 2018

Answer:

Re-arrange the equation, and divide through by #x#'s coefficient to find #x=5#

Explanation:

First, we'll re-arrange the equation such that all constants are on one side, and all of the #x#-related terms are on the other. We'll do this by adding and subtracting terms from both sides so terms will cancel out on one side, but modify the other:

First, we'll move the #4/5x# term:

#cancel(-4/5x)+6color(red)(cancel(+4/5x))=x-3color(red)(+4/5x)#

#6=x(1+color(red)(4/5))-3#

#6=9/5x-3#

Next, we'll move the -3 constant to the left-hand side:

#6color(blue)(+3)=9/5xcancel(-3)color(blue)(cancel(+3))#

#9=9/5x#

Now, all that's left is to divide both sides by #x#'s coefficient. Because we're dividing a fraction, we can write it as both sides being multiplied by its inverse:

#cancel(9)color(red)(xx5/cancel(9))=cancel(9/5)x xxcolor(red)(cancel(5/9))#

#color(green)(x=5)#