Answer:
The amount invested at 7% was $1,500 and the amount invested at 9% was $4,500
Step-by-step explanation:
we know that
The simple interest formula is equal to
[tex]I=P(rt)[/tex]
where
I is the Final Interest 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
Let
x-----> the amount invested at account paying 7%
(6,000-x) -----> the amount invested at account paying 9%
[tex]t=1\ years\\ P1=\$x\\ P2=\$(6,000-x)\\I=\$510\\r1=0.07\\r2=0.09[/tex]
substitute in the formula above
[tex]I=P1(r1t)+P2(r2t)[/tex]
[tex]510=x(0.07*1)+(6,000-x)(0.09*1)[/tex]
Solve for x
[tex]510=0.07x+540-0.09x[/tex]
[tex]0.09x-0.07x=540-510[/tex]
[tex]0.02x=30[/tex]
[tex]x=\$1,500[/tex]
[tex]6,000-x=6,000-1,500=\$4,500[/tex]
therefore
The amount invested at 7% was $1,500 and the amount invested at 9% was $4,500
