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

2 Answers
Jul 9, 2017

See a solution process below:

Explanation:

First, expand the terms on the left side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

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

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

#4x - 10 = 6x + 4#

Next, subtract #color(red)(4x)# and #color(blue)(4)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-color(red)(4x) + 4x - 10 - color(blue)(4) = -color(red)(4x) + 6x + 4 - color(blue)(4)#

#0 - 14 = (-color(red)(4) + 6)x + 0#

#-14 = 2x#

Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#-14/color(red)(2) = (2x)/color(red)(2)#

#-7 = (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2))#

#-7 = x#

#x = -7#

Jul 9, 2017

#x = -7#

Explanation:

First expand the brackets by multiplying each thing inside the brackets by 2:

#4x - 10 = 6x + 4#

then minus the #4x# so there are only #x#'s on one side:

#- 10 = 2x + 4#

then minus the #4# so there are only numbers on one side and #x#'s on the other:

#-14 = 2x#

then divide by #2# to get the value of #x#:

#-7 = x#