Answer:
$15,000 was invested in CDs.
$40,000 was invested in bonds.
$75,000 was invested in stocks.
Step-by-step explanation:
Let C be the amount invested for CDs, B be the amount invested for bonds, and S be the amount invested for stocks.
We know that the total amount invested must be $130,000 because that was the amount given to the Scholarship Fund. So:
![C+B+S=130000](https://tex.z-dn.net/?f=C%2BB%2BS%3D130000)
We know that CDs pay 2% or 0.02 interest; Bonds pay 2.9% or 0.029 interest; and Stocks pay 9.9% or 0.099 interest. In total, the investment earned $8,885. So:
![0.02C+0.029B+0.099S=8885](https://tex.z-dn.net/?f=0.02C%2B0.029B%2B0.099S%3D8885)
Let's remove the decimals by multiplying everything by 1000. This yields:
![20C+29B+99S=8885000](https://tex.z-dn.net/?f=20C%2B29B%2B99S%3D8885000)
Now, we know that $25000 more was invested in bonds than CDs. So:
![B=25000+C](https://tex.z-dn.net/?f=B%3D25000%2BC)
We have a triple system of equations. We can solve this using substitution. Let's substitute our third equation into the second equation. This yields:
![20C+29(25000+C)+99S=8885000](https://tex.z-dn.net/?f=20C%2B29%2825000%2BC%29%2B99S%3D8885000)
We want to isolate the C variable. Here, we removed the B but we still have an S.
So, to remove the S, we can go back to our first equation. We have:
![C+B+S=13000](https://tex.z-dn.net/?f=C%2BB%2BS%3D13000)
Subtract C and B from both sides:
![S=130000-C-B](https://tex.z-dn.net/?f=S%3D130000-C-B)
We can now substitute this for S:
![20C+29(25000+C)+99(130000-C-B)=8885000](https://tex.z-dn.net/?f=20C%2B29%2825000%2BC%29%2B99%28130000-C-B%29%3D8885000)
We got a B again. But, we already know what B is. Substitute:
![20C+29(25000+C)+99(130000-C-((25000)+C))=8885000](https://tex.z-dn.net/?f=20C%2B29%2825000%2BC%29%2B99%28130000-C-%28%2825000%29%2BC%29%29%3D8885000)
So, our equation is now in terms of C. Solve for C. Distribute on the far right:
![20C+29(25000+C)+99(130000-C-25000-C)=8885000](https://tex.z-dn.net/?f=20C%2B29%2825000%2BC%29%2B99%28130000-C-25000-C%29%3D8885000)
Combine like terms:
![20C+29(25000+C)+99(105000-2C)=8885000](https://tex.z-dn.net/?f=20C%2B29%2825000%2BC%29%2B99%28105000-2C%29%3D8885000)
Distributive Property:
![20C+725000+29C+10395000-198C=8885000](https://tex.z-dn.net/?f=20C%2B725000%2B29C%2B10395000-198C%3D8885000)
Combine like terms:
![(20C+29C-198C)+(725000+10395000)=8885000](https://tex.z-dn.net/?f=%2820C%2B29C-198C%29%2B%28725000%2B10395000%29%3D8885000)
Add:
![-149C+11120000=8885000](https://tex.z-dn.net/?f=-149C%2B11120000%3D8885000)
Subtract 1112000 from both sides:
![-149C=-2235000](https://tex.z-dn.net/?f=-149C%3D-2235000)
Divide both sides by -149:
![C=15000](https://tex.z-dn.net/?f=C%3D15000)
So, a total of $15,000 was invested in CDs.
Since $25,000 more was invested in bonds than in CDs, this means that a total of $25000+$15000 or $40,000 was invested in bonds.
To find out how much was invested in stocks, we can use our first equation again:
![C+B+S=130000](https://tex.z-dn.net/?f=C%2BB%2BS%3D130000)
Substitute 15000 for C and 40000 for B:
![15000+40000+S=130000](https://tex.z-dn.net/?f=15000%2B40000%2BS%3D130000)
Add:
![55000+S=130000](https://tex.z-dn.net/?f=55000%2BS%3D130000)
Subtract 55000 from both sides:
![S=\$75000](https://tex.z-dn.net/?f=S%3D%5C%2475000)
So, $15,000 was invested in CDs, $40,000 was invested in bonds, and $75,000 was invested in stocks.
And we're done!