Answer:
The amount invested at 6% was $7,200 and the amount invested at 9% was $2,800
Step-by-step explanation:
Let
x ----> the amount invested at 6%
10,000-x -----> the amount invested at 9%
we know that
The interest earned in one year by the amount invested at 6% plus the interest earned by the amount invested at 9% must be equal to $684
Remember that
[tex]6\%=6/100=0.06[/tex]
[tex]9\%=9/100=0.09[/tex]
so
The linear equation that represent this problem is
[tex]0.06x+0.09(10,000-x)=684[/tex]
solve for x
[tex]0.06x+900-0.09x=684[/tex]
[tex]0.09x-0.06x=900-684[/tex]
[tex]0.03x=216[/tex]
[tex]x=\$7,200[/tex]
[tex]\$10,000-x=\$10,000-\$7,200=\$2,800[/tex]
therefore
The amount invested at 6% was $7,200 and the amount invested at 9% was $2,800
