invest money at 9% and 49000 more than three times the amount at 10%. the total amount earned is 59500. how much did he invest in each?

Question

this is intermediate algebra

Answer 1

.09x + .10(49K+3x) = 59.5K (K=$1000)

.39x = 59.5 -4.9 = 54.6

x= 54.6/.39 = $140,000 invested at 9%

49+3(140) = 49+420= 469 = $469,000 invested at 10%

.09(140) + .10(469) = 12.6+ 46.9 = 59.5 = $59,500