Question 1011030
<pre>
{{{2x+5=3kx+5}}}

Subtract 5 from both sides

2x=3kx

Divide both sides by 3x, which can only be
done for non-zero values of x.

{{{2x/(3x)=(3kx)/(3x)}}}

{{{2/3=k}}}

That's no doubt the desired value of k, but we must continue
on to show that it is also true when x=0 and k={{{2/3}}}.

So we investigate to see if it also holds true for x=0:

When we replace k by {{{2/3}}} in 

{{{2x+5=3kx+5}}}

we get 

{{{2x+5=3(2/3)x+5}}}

and when we substitute x=0

{{{2(0)+5=3(2/3)(0)+5}}}

{{{0+5=0+5}}}

{{{5=5}}}

Now since we have shown that it also holds for x=0, 
it holds for ALL values of x.

[To have a rigorous proof we needed to show that
it also held for x=0]

Edwin</pre>