How do you solve #1/8x+3/2=3/4x-1#?

2 Answers
Jun 26, 2017

Answer:

See a solution process below:

Explanation:

First, multiply both sides of the equation by #color(red)(8)# to eliminate the fractions. #color(red)(8)# is the Lowest Common Denominator (LCD) of the three fractions:

#color(red)(8)(1/8x + 3/2) = color(red)(8)(3/4x - 1)#

#(color(red)(8) xx 1/8x) + (color(red)(8) xx 3/2) = (color(red)(8) xx 3/4x) - (color(red)(8) xx 1)#

#(cancel(color(red)(8)) xx 1/color(red)(cancel(color(black)(8)))x) + (cancel(color(red)(8))4 xx 3/color(red)(cancel(color(black)(2)))) = (cancel(color(red)(8))2 xx 3/color(red)(cancel(color(black)(4)))x) - 8#

#1x + 12 = 6x - 8#

Next, subtract #color(red)(1x)# and add #color(red)(8)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-color(red)(1x) + 1x + 12 + color(red)(8) = -color(red)(1x) + 6x - 8 + color(red)(8)#

#0 + 20 = (-color(red)(1) + 6)x - 0#

#20 = 5x#

Now, divide each side of the equation by #color(red)(5)# to solve for #x# while keeping the equation balanced:

#20/color(red)(5) = (5x)/color(red)(5)#

#4 = (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))#

#4 = x#

#x = 4#

Jun 26, 2017

Answer:

#color(orange)(x!=4# is a extraneous solution

Explanation:

#1/8x+3/2=3/4x-1#

multiply L.H.S AND R.H.S by #8#

#:.x+12=6x-8#

#:.x-6x=-8-12#

#:.-5x=-20#

multiply L.H.S AND R.H.S by# -1#

#:.5x=20#

#:.color(orange)(x=4#

substitute #color(orange)(x=4#

#:.1/8(color(orange)4)+3/2=3/4(color(orange)4)-1#

#:.4/8+3/2=12/4-1#

multiply L.H.S AND R.H.S by#8#

#:.32/8+24/2=24/4-8#

#:.4+12=6-8#

#:.color(orange)(16!=-2#

#:.color(orange)(x=4 # is a extraneous solution