# algebra

Sam invested \$2150, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was \$22 more than twice the income from the 6% investment. How much did he invest at each rate?

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1. if \$x at 6%, the rest (2150-x) is at 8%. So, now we have

.08(2150-x) = 22+2(.06x)

now just crank it out.

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posted by Steve

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