this guy invested $10,000, part at 3% and part at 4% simple interest for a period of 1 year. how much was invested into each account if the interest earned in the 4% account was $246 greater than the 3% account?
.04x = .03(10000-x)+246
x = amount at 4%
10,000 -x = amount at 3%
I = Prt
Interest for amount at 4% is
.04x since time is 1.
Interest at 3% is .03(10000-x)
the 4% interest is $246 more
.04x = .03(10000-x) +246
To solve this problem, we need to set up a system of equations based on the given information. Let's call the amount invested at 3% x, and the amount invested at 4% y.
According to the problem, the total amount invested is $10,000.
So, the first equation is:
x + y = 10,000.
Next, we need to consider the interest earned. The interest earned on the 3% account is x * 0.03, and the interest earned on the 4% account is y * 0.04. We know from the problem that the interest earned on the 4% account is $246 greater than the 3% account. Therefore, we can set up the second equation as follows:
(y * 0.04) - (x * 0.03) = 246.
Now we have a system of two equations:
x + y = 10,000,
(y * 0.04) - (x * 0.03) = 246.
To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution.
From the first equation, we can solve for x and express it in terms of y:
x = 10,000 - y.
Now, we substitute this expression for x in the second equation:
(y * 0.04) - ((10,000 - y) * 0.03) = 246.
Simplifying the equation, we get:
0.04y - 0.03(10,000 - y) = 246,
0.04y - 300 + 0.03y = 246,
0.07y - 300 = 246.
Adding 300 to both sides of the equation, we have:
0.07y = 546.
Dividing both sides by 0.07, we find:
y = 7,800.
Now, we can substitute the value of y back into the first equation to solve for x:
x + 7,800 = 10,000,
x = 10,000 - 7,800,
x = 2,200.
So, $2,200 was invested at 3% and $7,800 was invested at 4% in order to earn an interest of $246 more on the 4% account.