amount invested at 5% --- x
amount invested at 2.5% --- 14000-x
solve for x
.05x + .025(14000-x) = 540
amount invested at 2.5% --- 14000-x
solve for x
.05x + .025(14000-x) = 540
We know that the total amount invested is $14,000, so we have the equation:
x + y = 14,000 ----(Equation 1)
We also know that the interest earned after 1 year was $540, and we can use the formula for simple interest to set up another equation:
0.05x + 0.025y = 540 ----(Equation 2)
Now we can solve the system of equations to find the values of x and y.
To eliminate decimals, let's multiply both sides of Equation 2 by 100 to clear the decimal points:
5x + 2.5y = 54,000 ----(Equation 3)
Now we can solve the system of equations composed of Equation 1 and Equation 3. We'll use the method of substitution:
From Equation 1, we can rearrange it to express x in terms of y:
x = 14,000 - y
Substituting this value of x into Equation 3:
5(14,000 - y) + 2.5y = 54,000
70,000 - 5y + 2.5y = 54,000
-2.5y = 54,000 - 70,000
-2.5y = -16,000
Dividing both sides of the equation by -2.5:
y = -16,000 / -2.5
y = 6,400
Now substitute this value of y back into Equation 1 to solve for x:
x + 6,400 = 14,000
x = 14,000 - 6,400
x = 7,600
Therefore, the investment adviser invested $7,600 in the account that earned 5% interest and $6,400 in the account that earned 2.5% interest.
Let's say the amount invested at 5% interest is x dollars, and the amount invested at 2.5% interest is $14,000 - x dollars (since the total investment amount is $14,000).
The equation can be set up as follows:
0.05x + 0.025(14,000 - x) = 540
Let's simplify the equation:
0.05x + 0.025(14,000) - 0.025x = 540
0.05x + 350 - 0.025x = 540
0.025x + 350 = 540
Now, subtract 350 from both sides:
0.025x = 540 - 350
0.025x = 190
To isolate x, divide both sides by 0.025:
x = 190 / 0.025
x = 7600
So, $7,600 was invested at 5% interest.
To find the amount invested at 2.5% interest, subtract $7,600 from the total investment amount:
$14,000 - $7,600 = $6,400
Therefore, $6,400 was invested at 2.5% interest.
In summary, $7,600 was invested at 5% interest, and $6,400 was invested at 2.5% interest.