Question 101825
solving for x
{{{(2x+1)/5=(x-6)/2}}}
first move everything to one side of the equation
lets move {{{(x-6)/2}}} to the left side of the equation
to do this just subtract it from both sides like this:
{{{((2x+1)/5)-((x-6)/2)=((x-6)/2)-((x-6)/2)}}}
the right side cancels out so we are left with this
{{{((2x+1)/5)-((x-6)/2)=0}}}
Now to subtract the fractions on the left side we need find the LCD (least common denominator)The LCD for 5 and 2 is 10 so that gives us this

{{{((4x+2)/10)-((5x-30)/10)=0}}}
now we can subtract the fractions
{{{(-x+32)/10=0}}}
multiply both sides of the equation by 10
{{{-x+32=0}}}
subtract 32 from both sides
{{{-x+32-32=0-32}}}
{{{-x = -32}}}
divide both sides by -1
{{{x=32}}}
<b>Answer: x = 32</b>
Check:
replace x in original equation with 32
{{{(2x+1)/5=(x-6)/2}}}
{{{(2(32)+1)/5=(32-6)/2}}}
{{{(64+1)/5=26/2}}}
{{{65/5=26/2}}}
{{{13=13}}}

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here is another approach to solving this problem 
{{{(2x+1)/5=(x-6)/2}}}
multiply both sides by 5
{{{5*(2x+1)/5=5*(x-6)/2}}}
{{{2x+1=(5x-30)/2}}}
now multiply both sides by 2
{{{2(2x+1)=2*(5x-30)/2}}}
{{{4x+2=5x-30}}}
move all x terms to one side of the equation and 
all terms with out x to the opposite side of the equation
{{{30+2=5x-4x}}}
{{{32 = x}}}