Answer:
In the first account was invested [tex]\$3,000[/tex] at 3%
In the second account was invested [tex]\$4,000[/tex] at 5%
Step-by-step explanation:
we know that
The simple interest formula is equal to
[tex]I=P(rt)[/tex]
where
I is the 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 we have
First account
[tex]t=1 years\\ P=\$x\\ r=0.03[/tex]
substitute in the formula above
[tex]I=x(0.03*1)[/tex]
[tex]I=0.03x[/tex]
Second account
[tex]t=1 years\\ P=\$(7,000-x)\\ r=0.05[/tex]
substitute in the formula above
[tex]I=(7,000-x)(0.05*1)[/tex]
[tex]I=350-0.05x[/tex]
Remember that
The interest is equal to [tex]\$290[/tex]
so
Adds the interest of both accounts
[tex]0.03x+350-0.05x=\$290[/tex]
[tex]0.05x-0.03x=350-290[/tex]
[tex]0.02x=60[/tex]
[tex]x=\$3,000[/tex]
therefore
In the first account was invested [tex]\$3,000[/tex] at 3%
In the second account was invested [tex]\$7,000-\$3,000=\$4,000[/tex] at 5%
