How do you solve #\frac { 5x - 8} { 2} \geq \frac { 4x - 2} { 3}#?

1 Answer
Jun 11, 2017

See a solution process below:

Explanation:

First, multiply each side of the inequality by #color(red)(6)# to eliminate the factions while keeping the inequality balanced. #color(red)(6)# is the lowest common denominator of the two fractions:

#color(red)(6) xx (5x - 8)/2 >= color(red)(6) xx (4x - 2)/3#

#cancel(color(red)(6))3 xx (5x - 8)/color(red)(cancel(color(black)(2))) >= cancel(color(red)(6))2 xx (4x - 2)/color(red)(cancel(color(black)(3)))#

#3(5x - 8) >= 2(4x - 2)#

Next, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(3)(5x - 8) >= color(blue)(2)(4x - 2)#

#(color(red)(3) xx 5x) - (color(red)(3) xx 8) >= (color(blue)(2) xx 4x) - (color(blue)(2) xx 2)#

#15x - 24 >= 8x - 4#

Then, add #color(red)(24)# and subtract #color(blue)(8x)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#-color(blue)(8x) + 15x - 24 + color(red)(24) >= -color(blue)(8x) + 8x - 4 + color(red)(24)#

#(-color(blue)(8) + 15)x - 0 >= 0 + 20#

#7x >= 20#

Now, divide each side of the inequality by #color(red)(7)# to solve for #x# while keeping the inequality balanced:

#(7x)/color(red)(7) >= 20/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) >= 20/7#

#x >= 20/7#