How do you solve #-(x+3)+3/4x+5=0#?

1 Answer
Feb 12, 2017

Answer:

See the entire solution process below:

Explanation:

First, remove the terms in parenthesis being sure to handle the signs for the individual terms correctly:

#-x - 3 + 3/4x + 5 = 0#

Next, multiply each side of the equation by #color(red)(4)# to eliminate the fraction while keeping the equation balanced:

#color(red)(4)(-x - 3 + 3/4x + 5) = color(red)(4) xx 0#

#(color(red)(4) xx -x) - (color(red)(4) xx 3) + (color(red)(4) xx 3/4x) + (color(red)(4) xx 5) = 0#

#-4x - 12 + (cancel(color(red)(4)) xx 3/color(red)(cancel(color(black)(4)))x) + 20 = 0#

#-4x - 12 + 3x + 20 = 0#

Then, combine line terms:

#-4x + 3x - 12 + 20 = 0#

#(-4 + 3)x + 8 = 0#

#-x + 8 = 0#

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

#color(red)(x) - x + 8 = color(red)(x) + 0#

#0 + 8 = x#

#8 = x#

#x = 8#