How do you solve #1/2x-5/3=-1/2x+19/4#?

2 Answers
Jan 25, 2017

See the entire solution process below:

Explanation:

First, multiply both sides of the equation by #color(red)(12)# (the lowest common denominator of all the fractions) to eliminate the fractions while keeping the equation balanced:

#color(red)(12)(1/2x - 5/3) = color(red)(12)(-1/2 x + 19/4)#

#(color(red)(12) xx 1/2x) - (color(red)(12) xx 5/3) = (color(red)(12) xx -1/2x) + (color(red)(12) xx 19/4)#

#(6 xx 1x) - (4 xx 5) = (6 xx -1x) + (3 xx 19)#

#6x - 20 = -6x + 57#

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

#6x - 20 + color(red)(6x) + color(blue)(20) = -6x + 57 + color(red)(6x) + color(blue)(20)#

#6x + color(red)(6x) - 20 + color(blue)(20) = -6x + color(red)(6x) + 57 + color(blue)(20)#

#12x - 0 = 0 + 77#

#12x = 77#

Now, divide each side by #color(red)(12)# to solve for #x# while keeping the equation balanced:

#(12x)/color(red)(12) = 77/color(red)(12)#

#(color(red)(cancel(color(black)(12)))x)/cancel(color(red)(12)) = 77/12#

#x = 77/12#

Jan 25, 2017

#x = "77"/"12"#

Explanation:

#"1"/2"x" - "5"/"3" = -"1"/2"x" + "19"/"4"#

#"1"/2"x" + "1"/2"x" = "19"/"4" + "5"/"3"#

#"1x + 1x"/"2" = "(19)(3) + (5)(4)"/"12"#

#"2x"/"2" = "57 + 20"/"12"#

#x = "77"/12"#