Question 328981
A woman invested some money at 8% and some at 9%.  The interest on the combined investment of $10,000 was $840.  How much was invested at 9%

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let her invest $x at 8%
$y at 9%
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x+y=10000....................1
0.08x+0.09y=840
multiply by 100
8x+9y=84000..................2
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multiply equation 1 by 8 and subtract equation 2
8x+8y-8x-9y=80000-84000
-y=-4000
y= $4000 the amount investd at 9%
plug the value of y in equation 1
x+4000=10000
x= $6000 amount invested at8%