How do you solve #x/2 + x/3 - 1 = x/6 + 3#?

2 Answers
Leslie · Veddesh #phi#
Feb 2, 2016

#(x/2*3/3) + (x/3*2/2) - x/6 = 4 rarr (4x)/6 = 4#, therefore, #x = 6#

Explanation:

To add and subtract fractions they must have a common denominator. The common denominator for #2, 3# and #6# is #6#, therefore #x/2# is multiplied by #3/3# (any number over itself = #1#, so you are not changing the value of the fraction by multiplying it by #1#) and #x/3# is multiplied by #2/2#, resulting in #(3x)/6 + (2x)/6#, which equals #(5x)/6#. Now we move all of the fractions to one side of the equation and the whole numbers to the other:

#(5x)/6 - x/6 - 1 + 1 = x/6 - x/6 + 1 + 3#

#= (4x)/6 = 4#

#6 * 4 = 4x#, therefore, #x = 6#

Veddesh #phi#
Feb 2, 2016

#x/2+x/3-1=x/6+3#

#rarr(3x+2x)/6-1=x/6+3#

#rarr(5x)/6-1=x/6+3#

#rarr(5x)/6=x/6+3+1#

#rarr(5x)/6=x/6+4#

#rarr(5x)/6-x/6=4#

#rarr(4x)/6=4#

#rarr(2x)/3=4#

#rarr2x=4*3#

#rarr2x=12#

#rarrx=12/2=6#

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