Question 61422: John Craig invested a portion of $15,000 at 10% annual interest and the balance at 6% annual interest. If he earned $1260 for the year from the two accounts, how much did he invest at 10%?
Answer by tutorcecilia(2152)   (Show Source):
You can put this solution on YOUR website! Ax+By=C [use the standard form of a line]
10%x+6%y=$1260
.10x+.06y=1260 [convert the percents to decimals]
and
x+y=$15,000
x+($15,000-x)=$15,000
putting it all together gives:
.10x+.06(15,000-x)=1260 [substitute (15,000-x) for y]
.10x+900-.06=1260
.04x=360
x=$9000 invested at 10%
y=(15,000-x)=15,000-9000=$6,000 invested at 6%
because $6,000+$9,000=$15,000
and
.10(9000)=$900
.06(6,000)=$360
900+360=$1260
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