Question 153423
<pre><font size = 4 color = "indigo"><b>
{{{(1/8)(7m-7)=(2/5)(4m-4)+8}}}

Since two of the terms have fractions in front,
let's write {{{8}}} as the fraction {{{8/1}}}

{{{(1/8)(7m-7)=(2/5)(4m-4)+8/1}}}

The fractions have denominators 8, 5, and 1
so the LCD = 40 so we multiply through every
term by {{{40/1}}}


{{{(40/1)(1/8)(7m-7)=(40/1)(2/5)(4m-4)+(40/1)(8/1)}}}
 
 {{{5}}}            {{{8}}}        
{{{(cross(40)/1)(1/cross(8))(7m-7)=(cross(40)/1)(2/cross(5))(4m-4)+(40/1)(8/1)}}}

{{{5(7m-7)=16(4m-4)+320}}}

{{{35m-35=64m-64+320}}}

{{{35m-35=64m+256}}}

{{{-29m=291}}}

{{{m=(291)/(-29)}}}

{{{m=-291/29}}}

Edwin</pre>