PROBLEM
Systems of equation question
A person invests $35000, partly at 5.00%, partly at 6.00%, and the remainder at 6.50%. The total annual interest is $ 2,105. Four times the amount invested at 6.00% equals the amount invested at 5.00% and 6.50% combined. How much is invested at each rate?
A person invests $35000, partly at 5.00%, partly at 6.00%, and the remainder at 6.50%. The total annual interest is $ 2,105. Four times the amount invested at 6.00% equals the amount invested at 5.00% and 6.50% combined. How much is invested at each rate?
SOLUTION
Let X be the amount invested at 5%.
Let Y be the amount invested at 6%.
Let Z be the amount invested at 6.5%.
Therefore, the system of equations is as follows:
EQUATION 1 (INVESTMENT TOTAL): X + Y + Z = 35000
EQUATION 2 (INTEREST): 0.05X + 0.06Y + 0.065Z = 2015
EQUATION 3 (INVESTMENT RELATIONSHIPS): 4Y = X + Z
USE EQUATION 3 SUBSTITUTE THE VALUE OF X + Z INTO EQUATION 1 AND SOLVE FOR Y
Y + X + Z = 35000
Y + 4Y = 35000
5Y = 35000
Y = 7000
SUBSTITUTE VALUE OF Y INTO EQUATIONS 1 AND 2 AND SOLVE FOR X AND Z
X + 7000 + Z = 35000
X + Z = 28000 (EQUATION 1)
0.05X + 0.06(7000) + 0.065Z = 2105
0.05X + 420 + 0.065Z = 2105
0.05X +0.065Z = 1685 (EQUATION 2)
USING ELIMINATION METHOD
X + Z = 28000
0.05X + 0.065Z = 1685
MULTIPLYING EQUATION 1 BY -0.05
- 0.05X - 0.05Z = - 1400
0.05X + 0.065Z = 1685
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0.15Z = 285
Z = 19000
X + 19000 = 28000
X = 9000
Therefore:
$9,000 is the amount invested at 5%.
$7,000 is the amount invested at 6%.
$19,000 is the amount invested at 6.5%.
$7,000 is the amount invested at 6%.
$19,000 is the amount invested at 6.5%.
