Question 55136
Lets call x the amount that Bob invested at 14% and y the amount that he invested at 13%. 

Combined, Bob invested 20 grand, so we have
{{{x + y = 20000}}}

Additionally, we know that the sum of the amount that Bob invested at 14% and 13% is equal to $2720. In math-speak, that says:

{{{.14x+.13y = 2720}}}

So we have the following system of equations:

{{{x + y = 20000}}}
{{{.14x + .13y = 2720}}}

My strategy will be to use the substitution method. 

Starting with
{{{x + y = 20000}}} and solving for x, I get
{{{x = 20000-y}}}

I'll plop that into the second equation
{{{.14x + .13y = 2720}}}
{{{.14(20000-y) + .13y = 2720}}}
{{{2800-.14y+.13y = 2720}}}
{{{2800 - .01y = 2720}}}
{{{-.01y = -80}}}
{{{y =8000}}}

Great! Bob invested $8000 at 13%

I'll put that back into {{{x = 20000-y}}} to find x
{{{x = 20000-8000}}}
{{{x = 12000}}}

So the answer is Bob invested $12000 at 14%

CHECK:

I'll check my work by plugging x = 12000 and y = 8000 into the second equation
{{{.14x+.13y = 2720}}}
{{{.14(12000)+.13(8000) = 2720}}}
{{{1680 + 1040 = 2720}}}
{{{2720 = 2720}}}...check!