How do you solve #-5(4x-2)=-2(3+6x)#?

1 Answer
Jul 15, 2017

Answer:

See a solution process below:

Explanation:

First, expand the terms within the parenthesis on each side of the equation. Multiply each term within the parenthesis by the term outside the parenthesis. Be careful to manage the signs of the individual terms correctly:

#color(red)(-5)(4x - 2) = color(blue)(-2)(3 + 6x)#

#(color(red)(-5) xx 4x) - (color(red)(-5) xx 2) = (color(blue)(-2) xx 3) + (color(blue)(-2) xx 6x)#

#-20x - (-10) = -6 + (-12x)#

#-20x + 10 = -6 - 12x#

Next, add #color(red)(20x)# and #color(blue)(6)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#-20x + 10 + color(red)(20x) + color(blue)(6) = -6 - 12x + color(red)(20x) + color(blue)(6)#

#-20x + color(red)(20x) + 10 + color(blue)(6) = -6 + color(blue)(6) - 12x + color(red)(20x)#

#0 + 16 = 0 + (-12 + color(red)(20))x#

#16 = 8x#

Now, divide each side of the equation by #color(red)(8)# to solve for #x# while keeping the equation balanced:

#16/color(red)(8) = (8x)/color(red)(8)#

#2 = (color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8))#

#2 = x#

#x = 2#