Peter has money in two savings accounts. One rate is 9% and the other is 11%. If he has $400 more in the 11% account and the total interest is $117, how much is invested in each savings account?
.09x + .11(x+400) = 117.00
Show me the steps pleeeeeeeeeeeeeeeeeeeeeease.
Let's assume that the amount invested in the 9% account is x dollars.
According to the information given, the amount invested in the 11% account is $400 more than the amount in the 9% account. So, the amount invested in the 11% account would be (x + $400) dollars.
Now, let's calculate the interest earned from each account:
Interest from the 9% account = x * 9% = 0.09x dollars
Interest from the 11% account = (x + $400) * 11% = 0.11(x + 400) dollars
The total interest earned is given as $117, so we can set up the equation:
0.09x + 0.11(x + 400) = $117
Now, let's solve for x:
0.09x + 0.11x + 44 = $117
0.20x = $73
x = $73 / 0.20
x ≈ $365
Therefore, Peter has invested approximately $365 in the 9% account and $365 + $400 = $765 in the 11% account.
To find out how much is invested in each savings account, let's assign some variables. Let "x" be the amount invested in the 9% account and "y" be the amount invested in the 11% account.
We know that Peter has $400 more in the 11% account, so we can set up the equation: y = x + $400.
The total interest earned is given as $117. The interest earned from the 9% account can be calculated using the formula: 0.09x, and the interest earned from the 11% account is: 0.11y.
So, we can set up the equation: 0.09x + 0.11y = $117.
Now we can substitute the value of y from the first equation into the second equation: 0.09x + 0.11(x + $400) = $117.
Simplifying this equation will help us solve for x.
0.09x + 0.11x + $44 = $117
0.20x = $73
x = $73 / 0.20
x = $365
Now that we have the value of x, we can substitute it back into the first equation to find y.
y = x + $400
y = $365 + $400
y = $765
Therefore, Peter has $365 invested in the 9% account and $765 in the 11% account.