Question 128130
1.
{{{3/2}}} + {{{3/x}}}
------- 
2 + {{{4/x}}}
:
Put the fractions in the numerator & denominator over a single common denominator
{{{((3x + 2(3)))/(2x)}}}
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{{{((2x + 4))/x}}}
:
{{{((3x + 6))/(2x)}}}
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{{{((2x + 4))/x}}}
:
Factor out 3 in the numerator & 2 in the denominator
{{{(3(x + 2))/(2x)}}}
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{{{(2(x + 2))/x}}}
;
Remember when you divide fractions, invert the dividing fraction and multiply
{{{(3(x+2))/(2x)}}} * {{{x/(2(x+2))}}}
:
Cancel out (x+2) and x, leaving
{{{3/2}}} * {{{1/2}}} = {{{3/4}}}
:
:
2.
This is the same as the first one
:
3. This is similar to the first equation after we put it over common denominators
{{{((3x+6))/(2x)}}} 
----------
{{{((2x+4))/x}}}
:
Factor out 3 in the numerator & 2 in the denominator
{{{(3(x + 2))/(2x)}}}
----------
{{{(2(x + 2))/x}}}
;
Remember when you divide fractions, invert the dividing fraction and multiply
{{{(3(x+2))/(2x)}}} * {{{x/(2(x+2))}}}
:
Cancel out (x+2) and x, leaving
{{{3/2}}} * {{{1/2}}} = {{{3/4}}}
:
:
5. 
6x^2+24x+24
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2x^2 
:
notice that you can factor out 6 in the numerator, then cancel 2 into 6
{{{(6(x^2 + 4x + 4))/(2x^2)}}} = {{{(3(x^2 + 4x + 4))/(x^2)}}}
:
The numerator has a perfect square and can be factored to:
{{{(3(x+2)(x+2))/(x^2)}}} = {{{(3(x+2)^2)/(x^2)}}}