Question 1038878
<font size=3 color=black>

x = amount invested at 4%
y = amount invested at 3%



<font color=blue>"A man has $800 more invested at 4% than at 3%"</font> so we know that {{{x=y+800}}}


The amount of interest earned, in one year, for the 4% account is {{{x*0.04 = 0.04x}}}
The amount of interest earned, in one year, for the 3% account is {{{y*0.03 = 0.03y}}}


Put together, the total interest is {{{0.04x+0.03y}}} which is set equal to 130 because this is the total interest for the year.


We have {{{0.04x+0.03y=130}}}


---------------------------------------


Use the two equations {{{0.04x+0.03y=130}}} and {{{x=y+800}}} to solve for x and y. Start by using substitution. 



{{{0.04x+0.03y=130}}}


{{{0.04(y+800)+0.03y=130}}} Plug in {{{x=y+800}}}. Solve for y


{{{0.04y+0.04*(800)+0.03y=130}}}


{{{0.04y+32+0.03y=130}}}


{{{0.07y+32=130}}}


{{{0.07y+32-32=130-32}}}


{{{0.07y=98}}}


{{{0.07y/0.07=98/0.07}}}


{{{y=1400}}}


Now that we know that {{{y=1400}}}, use this to find x.


{{{x=y+800}}}


{{{x=1400+800}}}


{{{x=2200}}}



So we know that {{{x=2200}}} and {{{y=1400}}}



-------------------------------------


Final Answer: 


<font color=red>$2,200</font> was invested at 4%
<font color=red>$1,400</font> was invested at 3%

</font>