How do you solve #1/6-x/2=(x-5)/3#?

2 Answers
Jan 27, 2017

Answer:

see the entire solution process below:

Explanation:

First, multiple each side of the equation by #color(red)(6)# (the lowest common denominator of all the fractions) to eliminate the fractions while keeping the equation balanced:

#color(red)(6)(1/6 - x/2) = color(red)(6) xx (x - 5)/3#

#(color(red)(6) xx 1/6) - (color(red)(6) xx x/2) = cancel(color(red)(6))2 xx (x - 5)/color(red)(cancel(color(black)(3)))#

#(cancel(color(red)(6)) xx 1/color(red)(cancel(color(black)(6)))) - (cancel(color(red)(6))3 xx x/color(red)(cancel(color(black)(2)))) = 2(x - 5)#

#1 - 3x = 2x - 10#

Next, add #color(red)(3x)# and #color(blue)(10)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#1 - 3x + color(red)(3x) + color(blue)(10) = 2x - 10 + color(red)(3x) + color(blue)(10)#

#1 + color(blue)(10) - 3x + color(red)(3x) = 2x + color(red)(3x) - 10 + color(blue)(10)#

#11 - 0 = 5x - 0#

#11 = 5x#

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

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

#11/5 = (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))#

#11/5 = x#

#x = 11/5#

Jan 27, 2017

Answer:

#x=11/5#

Explanation:

Eliminate the fractions in the equation by multiplying ALL terms on both sides of the equation by the #color(blue)"lowest common multiple"# ( LCM ) of the denominators 6, 2 and 3

The LCM of 6 , 2 and 3 is 6

Hence multiply All terms by 6

#(cancel(6)^1xx1/cancel(6)^1)-(cancel(6)^3xx x/cancel(2)^1)=(cancel(6)^2xx(x-5)/cancel(3)^1)#

#rArr1-3x=2(x-5)larr" no fractions"#

distribute the bracket on the right side.

#rArr1-3x=2x-10#

subtract 2x from both sides.

#1-3x-2x=cancel(2x)cancel(-2x)-10#

#rArr1-5x=-10#

subtract 1 from both sides.

#cancel(1)cancel(-1)-5x=-10-1#

#rArr-5x=-11#

To solve for x, divide both sides by - 5

#(cancel(-5) x)/cancel(-5)=(-11)/(-5)#

#rArrx=11/5#

#color(blue)"As a check"#

substitute this value into the equation and if the left side equals the right side then it is the solution.

#"left side "=1/6-(11/5)/2=1/6-11/10=5/30-33/30=-28/30=-14/15#

#"right side"=(11/5-5)/3=(11/5-25/5)/3=(-14/5)/3=-14/15#

#rArrx=11/5" is the solution"#