Question 295568
<pre>
You must have meant {{{3*b}}} not {{{3^b}}}

because there is no solution to 

{{{(3^b)/5-1/5=b/5}}}

and similar equations cannot be solved except 
by iterative methods, not by the mothods of 
ordinary algebra. 

So please learn that {{{"3^b"}}} means {{{3^b}}}, not {{{3*b}}}

I will assume you meant:

{{{(3*b)/5-1/5=b/5}}}

Enclose every term in parentheses:

{{{((3*b)/5)-(1/5)=(b/5)}}}

The LCD is {{{red(5)}}}, so write it as {{{red((5/1))}}}
and multiply every term by that

{{{red((5/1))((3*b)/5)-red((5/1))(1/5)=red((5/1))(b/5)}}}  

Now cancel away the 5 denominators:

{{{red((cross(5)/1))((3*b)/cross(5))-red((cross(5)/1))(1/cross(5))=red((cross(5)/1))(b/cross(5))}}}

{{{3*b-1=b}}}

Add 1 to both sides

{{{3*b=b+1}}}

Add -b to both sides:

{{{2*b=1}}}

Divide both sides by {{{red(2)}}}

{{{(2*b)/red(2)=1/red(2)}}}

{{{(cross(2)*b)/red(cross(2))=1/red(2)}}}

{{{b=1/2}}}

Edwin</pre>