Question 686855
$8000 was invested at 15% and 
$5000 was invested at -13%.
The way to solve it depends on what you are studying
 
IF YOU ARE STUDYING SYSTEMS OF LINEAR EQUATIONS:
{{{x}}} = amount (in $) invested at 15%
{{{y}}} = amount (in $) invested at -13%
{{{highlight(x+y=13000)}}}
The first account earned ${{{0.15x}}} (15% of ${{{x}}})
The second account "earned" a negative amount (lost 13%),
so the "earnings" from the second account (in $) were {{{-0.13y}}}.
The total earnings (in $) were
{{{highlight(0.15x-0.13y=550)}}}
Putting both equations together, you get the system of equations
{{{system(x+y=13000,0.15x-0.13y=550)}}}
There are many ways to solve such a system,
but the most intuitive way would be by substitution.
You could solve the first equation for {{{y}}}:
{{{x+y=13000}}} --> {{{y=13000-x}}},
and then substitute the expression {{{13000-x}}} for {{{y}}} in the second equation.
{{{0.15x-0.13(13000-x)=550}}} --> {{{0.15x-0.13*13000+0.13x=550}}} --> {{{0.15x-1690+0.13x=550}}} --> {{{0.28x-1690=550}}} --> {{{0.28x=550+1690}}} --> {{{0.28x=2240}}} --> {{{x=2240/0.28}}} --> {{{highlight(x=8000)}}}
Then substitute {{{8000}}} for {{{x}}} in {{{y=13000-x}}} to get
{{{y=13000-8000}}} --> {{{highlight(y=5000)}}}
 
IF YOU HAVE NOT STUDIED SYSTEMS OF EQUATIONS:
{{{x}}} = amount (in $) invested at 15%
amount (in $) invested at -13% = {{{13000-x}}}
The first account earned ${{{0.15x}}} (15% of ${{{x}}})
The second account "earned" a negative amount (lost 13%),
so the "earnings" from the second account (in $) were {{{-0.13(13000-x)}}}.
The total earnings (in $) were
{{{highlight(0.15x-0.13(13000-x)=550)}}}
Now you simplify and solve for {{{x}}}
{{{0.15x-0.13(13000-x)=550}}} --> {{{0.15x-0.13*13000+0.13x=550}}} --> {{{0.15x-1690+0.13x=550}}} --> {{{0.28x-1690=550}}} --> {{{0.28x=550+1690}}} --> {{{0.28x=2240}}} --> {{{x=2240/0.28}}} --> {{{highlight(x=8000)}}}
Then,
amount (in $) invested at -13% = {{{13000-x=13000-800=highlight(5000)}}}