SOLUTION: John invested a portion of $15,000 at 10% annual interest and the rest at 6% annual interest. If he earned $1,260 for the year from the two accounts, how much did he invest at 10%?

Question 537454: John invested a portion of $15,000 at 10% annual interest and the rest at 6% annual interest. If he earned $1,260 for the year from the two accounts, how much did he invest at 10%?
Found 2 solutions by mananth, lmeeks54:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Investment I 6.00% per annum ---x
Investment II 10.00% per annum ---y

x + y= 15000 ------------------------1
6.00% x + 10.00% y= = $1,260.00
Multiply by 100
6 x + 10 y= = $126,000.00 --------2

Multiply (1) by -6
we get

-6 x -6 y= = -90000.00

Add this to (2)

0 x 4 y= = $36,000.00

divide by 4

y = $9,000.00 investment at 10.00%
Balance $6,000.00 investment at 6.00%
CHECK
$6,000.00 --------- 6.00% ------- $360.00
$9,000.00 ------- 10.00% ------- $900.00
Total -------------- $1,260.00
m.ananth@hotmail.ca
Answer by lmeeks54(111) (Show Source): You can put this solution on YOUR website!
Let H = # $ invested at 10%
Let L = # $ invested at 6%
...
Given:
H + L = 15,000
.10H + .06L = 1,260
...
With a system of two equations with two unknowns, rewrite one of the equations so that it is written for one unknown in terms of the other unknown.
...
H + L = 15,000
H = 15,000 - L
...
substitute H = 15,000 - L into the 2nd equation:
.10(15,000 - L) + .06L = 1,260
...
simplify by combining like terms:
1,500 - .10L + .06L = 1,260
...
subtract 1,500 from both sides:
-.04L = -240
...
divide both sides by -.04:
L = $6,000
...
This represents the amount of money invested at 6%. To find the amount of money invested at 10%, go back to our H = 15,000 - L equality and solve for H:
...
H = 15,000 - 6,000
H = $9,000
...
H = $9,000 equals the amount of money invested at 10%.
...
Check your work by inserting H = 9,000 and L = 6,000 into the 2nd equation:
.10(9,000) + .06(6,000) = 1,260
900 + 360 = 1,260
1,260 = 1,260 checks...
...
Cheers,
Lee
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