How do you solve #3/5(15x+20)+1/6(18x-12)=38#?

1 Answer
Jan 6, 2017

Answer:

See the full process below in the Explanation

Explanation:

Step 1) Expand the terms within the parenthesis:

#color(red)(3/5)(15x + 20) + color(blue)(1/6)(18x - 12) = 38#

#(color(red)(3/5) xx 15x) + (color(red)(3/5)xx 20) + (color(blue)(1/6) xx 18x) - (color(blue)(1/6) xx 12) = 38#

#(color(red)(3/cancel(5)) xx color(red)(cancel(color(black)(15)))3x) + (color(red)(3/cancel(5))xx color(black)(cancel(color(red)(20)))4) + (color(blue)(1/cancel(6)) xx color(blue)(cancel(color(black)(18)))3x) - (color(blue)(1/cancel(6)) xx color(blue)(cancel(color(black)(12)))2) = 38#

#(color(red)(3) xx 3x) + (color(red)(3)xx 4) + (color(blue)(1) xx 3x) - (color(blue)(1) xx 2) = 38#

#9x + 12 + 3x - 2 = 38#

We can now group and consolidate like terms:

#9x + 3x + 12 - 2 = 38#

#(9 + 3)x + 10 = 38#

#12x + 10 = 38#

We can next subtract #color(red)(10)# form each side of the equation to isolate the #x# term while keeping the equation balanced:

#12x + 10 - color(red)(10) = 38 - color(red)(10)#

#12x + 0 = 28#

#12x = 28#

Now we can divide each side of the equation by #color(red)(12)# to solve for #x# and keep the equation balanced:

#(12x)/color(red)(12) = 28/color(red)(12)#

#(color(black)(cancel(color(red)(12)))x)/cancel(color(red)(12)) = (4 xx 7)/(4 xx 3)#

#x = (cancel(4) xx 7)/(cancel(4) xx 3)#

#x = 7/3#