Question 84209
3. Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of $130. Randy also invested in the same scheme, but he interchanged the amounts invested, and received $4 more as interest. How much amount did each of them invest at different rates?

Here is how you should approach this problem
Say she invested x dollars at 12%, to represent % as a number divide % by 100, so 12% is .12. Say she invested y dollars at 10%, so .10. From this she earned 130 dollars. So you have your first equation
.12x+.10y=130
Next if you interchange amounts invested and add 4 dollars to earnings you have sencond equation
.10x+.12y=134
so use substitution to solve a system of equations

(1)   .12x+.10y=130
(2)   .10x+.12y=134

so, from (2) .1x+.12y=134
{{{x=(134-.12y)/.1}}}
(3) {{{x=1340-1.2y}}} now sub this in (1) in place of x, you get
{{{.12(1340-1.2y)+.1y=130}}}
{{{160.8-.144y+.1y=130}}}
{{{-.044y=-30.8}}}
{{{y=700}}} now plug this value into (3), you get
{{{x=1340-1.2(700)}}}
{{{x=1340-840}}}
{{{x=500}}}


So JANE invested $500 at 12% and $700 at 10% and
RANDY invested $700 at 12% and $500 at 10% 
Good luck