$6300 is invested,part of it at 10% and part of it at 7%.For a certain year, the total yield is $534.00. How much was invested at each rate
amount invested at 10% --- x
amount invested at 7% --- 6300-x
solve for x ....
.1x + .07(6300-x) = 534
I suggest multiplying each side by 100
$6300 is invested, part of it at 10% and part of it at 69%. For a certain year the total yield is $598.00. How much invested at each rate?
To find out how much was invested at each rate, let's set up two equations based on the given information.
Let's assume x represents the amount that was invested at 10% and y represents the amount that was invested at 7%.
Since the total amount invested is $6300, we know that x + y = 6300. Equation 1.
We are also given that the total yield from the investments is $534. The yield from the amount invested at 10% can be calculated as 0.10x, and the yield from the amount invested at 7% can be calculated as 0.07y. The sum of these two yields should be equal to $534.
So, we have the equation 0.10x + 0.07y = 534. Equation 2.
Now, we can solve the system of equations (equations 1 and 2) to find the values of x and y.
Using substitution or elimination method, let's solve for x:
From equation 1: x = 6300 - y.
Substitute this value of x into equation 2:
0.10(6300 - y) + 0.07y = 534.
Distribute 0.10:
630 - 0.10y + 0.07y = 534.
Combine like terms:
-0.03y = 534 - 630,
-0.03y = -96.
Multiply both sides by -1/0.03 to solve for y:
y = (-96)/(-0.03).
Calculating y:
y ≈ 3200.
Now, substitute the value of y back into equation 1 to find x:
x = 6300 - 3200,
x ≈ 3100.
So, approximately $3100 was invested at 10% and $3200 was invested at 7%.