# math

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A person has invested \$6000 . Part of the money is invested at 3% and the remaining amout at 4% . The annual income from the two investment is \$225.How much is invested at each rate?

• math -

amount invested at 3% ---- x
amount invested at 4% --- 6000-x

solve for x
.03x + .04(6000-x) = 225

I would multiply each of the 3 terms by 100 to get
3x + 4(6000-x) = 22500

I will let you finish it

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