Respuesta :
Let x be the money used in the bank account that pays 4% and let y be the money used for the 10% certificates. We have that 1/3x is the money used to buy the bonds. Since the total investment is $10,000, then we have the following equation:
[tex]\begin{gathered} x+\frac{1}{3}x+y=10000 \\ \Rightarrow\frac{4}{3}x+y=10000 \\ \text{ multiply by 3 both sides:} \\ (\frac{4}{3}x+y=10000)\cdot3=4x+3y=30000 \end{gathered}[/tex]to find the return of each investment, we change the rate to decimals and multiply it to the amount of the investment:
[tex]\begin{gathered} 0.04x\colon\text{ bank money} \\ 0.05(\frac{1}{3}x)\colon\text{ bonds money} \\ 0.1y\colon\text{ certificate deposits} \end{gathered}[/tex]since the total return of the investment is $885, then we have the following equation:
[tex]\begin{gathered} 0.04x+0.05(\frac{1}{3}x)+0.1y=885 \\ \text{ multiplying both sides by 3:} \\ (0.04x+0.05(\frac{1}{3}x)+0.1y=885)\cdot3=0.12x+0.05x+0.3y=2655 \\ \Rightarrow0.17x+0.3y=2655 \\ \text{ multiplying both sides by 10:} \\ (0.17x+0.3y=2655)\cdot10=1.7x+3y=26550 \end{gathered}[/tex]Now we have the following system of equations:
[tex]\begin{gathered} 4x+3y=30000 \\ 1.7x+3y=26550 \end{gathered}[/tex]We can substract the second equation from the first to get the following:
[tex]\begin{gathered} 4x+3y=30000 \\ -(1.7x+3y=26550) \\ ------------ \\ 2.3x=3450 \\ \Rightarrow x=\frac{3450}{2.3}=1500 \\ x=1500 \end{gathered}[/tex]we have that x=1500. This means that the money on the bankn account is $1500. Now we can use this value to find the money for the bonds and the money for the certificates:
[tex]\begin{gathered} x=1500 \\ \text{ money for the bonds:} \\ \frac{1}{3}x=\frac{1}{3}(1500)=500 \\ \text{money for the certificates:} \\ 4x+3y=30000 \\ \Rightarrow4(1500)+3y=30000 \\ \Rightarrow3y=30000-6000=24000 \\ \Rightarrow y=\frac{24000}{3}=8000 \\ y=8000 \end{gathered}[/tex]finally, we have that the money used for the bonds is $500 and for the certificates is $8000.
Now, suppose that the controller don't buy the bonds. Then the equations would be the following:
[tex]\begin{gathered} x+y=10000 \\ 0.04x+0.1y=885 \end{gathered}[/tex]Solving the system we get the following:
[tex]\begin{gathered} \text{first multiply the second equation by 10:} \\ (0.04x+0.1y=885)\cdot10 \\ \Rightarrow0.4x+y=8850 \\ \text{then we substract the second equation from the first:} \\ x+y=10000 \\ -(0.4x+y=8850) \\ ------------- \\ 0.6x=1150 \\ \Rightarrow x=\frac{1150}{0.6}=1916.67 \\ x=1916.67 \\ \Rightarrow1916.67+y=10000 \\ \Rightarrow y=10000-1916.67=8083.33 \\ y=8083.33 \end{gathered}[/tex]We can see that the solution changes with an increment on both quantities. Since the investments only were two, the money invested had to go up.
