How do you solve $x/5 - x/2= 3 + (3x)/10$ ?

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Answer 1 · 2016-07-08T14:57:42

$x= -5$

To solve $x/5 - x/2 = 3 + (3x)/10$

Begin by multiplying every term by the common denominator $(10)$
to eliminate the denominators

$(10)x/5 - (10)x/2 = (10)3 + (10)(3x)/10$

$cancel(10)2x/cancel5 - cancel(10)5x/cancel2 = (10)3 + cancel(10)(3x)/cancel(10)$

$2x -5x =30+3x$

Combine like terms

$-3x =30+3x$

Use the additive inverse to combine the variables on one side

$-3x -3x =30 cancel(+3x) cancel (-3x)$

$-6x = 30$

Use the multiplicative inverse to isolate the variable

$cancel(-6)x/cancel(-6) = 30/-6$

$x= -5$

How do you solve x/5 - x/2= 3 + (3x)/10x/5 - x/2= 3 + (3x)/10?