Investments- Mike perez has a total of $2000 on deposit with two savings institutions. One pays interest at a rate of 6% /year, the other pays interest at a rate of 8%/ year. If Mike earned a total of $144 in interest during a single year, how much d?

If Mike invested it all at 6%, his earnings would have been .06*$2000 = $120. He earned $24 more than that. His additional earnings were at an additional rate of 8% - 6% = 2%, so he must have had $24/.02 = $1200 invested at 8%.  Mike's investment was $1200 at 8% and $800 at 6%.

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Check   .08*1200 + .06*800 = 96 + 48 = 144  If you want to solve this algebraically, you can let "h" stand for the investment at the high interest rate.   .08h + .06(2000-h) = 144   .08h + 120 - .06h = 144    (use the distributive property to eliminate parentheses)   .02h + 120 = 144    (collect terms)   .02h = 24    (subtract 120)   h = 24/.02 = 1200    (if you read the above, this should look familiar)