Answer:
In an account at 2% invest [tex]\$4,800[/tex]
In an account at 8% invest [tex]\$1,200[/tex]
Step-by-step explanation:
we know that
The simple interest formula is equal to
[tex]A=P(1+rt)[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
in this problem we have two cases
First case
[tex]t=1\ year\\ P=\$x\\ A=\$6,000+\$192=\$6,192\\r=0.02[/tex]
Second case
[tex]t=1\ year\\ P=\$6,000-\$x\\ A=\$6,000+\$192=\$6,192\\r=0.08[/tex]
substitute both cases in the formula above
[tex]\$6,192=\$x(1+(0.02*1))+(\$6,000-\$x)(1+(0.08*1))[/tex]
solve for x
[tex]6,192=x(1.02)+(6,000-x)(1.08)[/tex]
[tex]6,192=1.02x+6,480-1.08x[/tex]
[tex]1.08x-1.02x=6,480-6,192[/tex]
[tex]0.06x=288[/tex]
[tex]x=\$4,800[/tex]
so
In an account at 2% invest [tex]\$4,800[/tex]
In an account at 8% invest [tex]\$1,200[/tex]
