Question 152443
{{{3/4}}}({{{4/5}}}{{{x-2}}})={{{11/4}}}
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First we get rid of the denominators of {{{4}}}
by multiplying both sides by {{{4/1}}}

{{{4/1}}}·{{{3/4}}}({{{4/5}}}{{{x-2}}})={{{4/1}}}·{{{11/4}}}

Cancel the {{{4}}}'s

{{{cross(4)/1}}}·{{{3/cross(4)}}}({{{4/5}}}{{{x-2}}})={{{cross(4)/1}}}·{{{11/cross(4)}}}

{{{3}}}({{{4/5}}}{{{x-2}}})={{{11}}}

Now write the {{{3}}} as {{{3/1}}}
and the {{{2}}} as {{{2/1}}}

{{{3/1}}}({{{4/5}}}{{{x-2/1}}})={{{11}}}

Now distribute on the left:

{{{3/1}}}·{{{4/5}}}{{{x}}}-{{{3/1}}}·{{{2/1=11}}}

{{{12/5}}}{{{x}}}-{{{6/1=11}}}

Write the {{{11}}} as {{{11/1}}}

{{{12/5}}}{{{x}}}-{{{6/1=11/1}}}

The LCD = {{{5}}} so we get rid of
the denominators by multiplying through
every term by {{{5/1}}}

{{{5/1}}}·{{{12/5}}}{{{x}}}-{{{5/1}}}·{{{6/1}}}={{{5/1}}}{{{11/1}}}

Now we cancel the 5's in the first term and
multiply in the other terms:

{{{cross(5)/1}}}·{{{12/cross(5)}}}{{{x}}}-{{{30/1}}}={{{55/1}}}

So all that's left is

{{{12x-30=55}}}

Add {{{30}}} to both sides:

{{{12x=85}}}

Divide both sides by {{{12}}}

{{{12x/12=85/12}}}

Cancel:

{{{cross(12)x/cross(12)=85/12}}}

{{{x=85/12}}}

Edwin</pre>