Problem 33 - College Algebra

Question

A person invested \(\$ 6700\) for one year, part at \(8 \%,\) part at \(10 \%,\) and the remainder at \(12 \% .\) The total annual income from these investments was \(\$ 716 .\) The amount of money invested at \(12 \%\) was \(\$ 300\) more than the amount invested at \(8 \%\) and \(10 \%\) combined. Find the amount invested at each rate.

Step-by-Step Solution

Answer

The amount invested at 8%, 10%, and 12% are \$2200, \$1000, and \$3500 respectively.

Step 1: Formulate the Equations

Let's designate three variables. Use \(x\) for the amount invested at 8%, \(y\) as the amount at 10%, and \(z\) as the invested amount at 12%. The three equations you would derive from the problem are: \[x + y + z = 6700\], \[0.08x + 0.10y + 0.12z = 716\] and \[z = 300 + x + y\].

Step 2: Substitute Third Equation

Substitute the \(z\) from the third equation into the first and second equations. This forms a new system of two equations: \[x + y + 300 + x + y = 6700\] which simplifies to \[2x + 2y = 6400\] and \[0.08x + 0.10y + 0.12(300 + x + y) = 716\] which simplifies to \[0.20x + 0.22y = 676\].

Step 3: Solve This New System of Equations

Divide the first equation by 2 to simplify into \(x + y = 3200\). From this deduced equation, express \(y = 3200 - x\). Substitute \(y\) into the other equation, which simplifies to: \[0.20x + 0.22(3200 - x) = 676\].

Step 4: Solve for x

Solve the equation: \[0.20x + 0.22(3200 - x) = 676\]. Which simplifies to: \[320 - 0.02x = 676\] and then solve for \(x\). \(x\) equals \(2200\).

Step 5: Solve for y and z

Substitute \(x = 2200\) into the equation \(y = 3200 - x\), we find that \(y\) equals \(1000\). Using \(z = 300 + x + y\), we find \(z\) equals \(3500\).

Step 6: Conclusion

Therefore, the amount invested at 8% was $2200, at 10% was $1000 and at 12% was $3500.