Question 199942
<font size = 8 color = "red"><b>Ray's solution is incorrect.
Here is the correct solution by Edwin:</b></font>

 {{{(5x+1)/7 - (2x-6)/4 >-4}}}
<pre><font size =4 color = "indigo"><b>

We will clear of fractions.

Put parentheses around each term and write the 
{{{-4}}} on the right as {{{-4/1}}} so
that every term will be a fraction:

 {{{((5x+1)/7) - ((2x-6)/4) >(-4/1)}}}

The LCD of all the denominators is {{{28}}}
Multiply every term through by {{{28/1}}}

 {{{(28/1)((5x+1)/7) - (28/1)((2x-6)/4) >(28/1)(-4/1)}}}

Cancel the 7 into the 28, getting 4

 {{{((""^4cross(28))/1    )((5x+1)/cross(7)) - (28/1)((2x-6)/4) >(28/1)(-4/1)}}}

Cancel the 4 into the 28 in the second term, getting 7

 {{{((""^4cross(28))/1    )((5x+1)/(cross(7))[1]) - ((""^7cross(28))/1)((2x-6)/(cross(4))[1]) >(28/1)(-4/1)}}}

Multiply the {{{28}}} by the {{{-4}}} on the right getting {{{-112}}}

 {{{((""^4cross(28))/1    )((5x+1)/(cross(7))[1]) - ((""^7cross(28))/1)((2x-6)/(cross(4))[1]) >-112}}}

What's left is

{{{4(5x+1) - 7(2x-6) > -112}}}

The we remove the parentheses by distributing:

{{{20x+4 - 14x+42 > -112}}}

{{{6x+46>-112}}}

{{{6x>-112-46}}}

{{{6x>-158}}}

{{{x>-158/6}}}

{{{x>-79/3}}}

The solution set in interval notation is ({{{-79/3}}},{{{infinity}}})

Edwin</pre>