Math Problem, College Algebra, Investing?

Spencer Heidtbrink: Let x be the amount invested at 4% and y be the amount invested at 3%x + y = 8000 or .or y = 8000 - x0.04x + 0.03y = 290substitute first equation into second equation:0.04x + 0.03(8000 - y) = 2900.04x + 240 - 0.03y = 2900.01x = 50x = 50/0.01 = 5000using first equation:y = 8000 - 5000 = 3000check using second equation:0.04(5000) + 0.03(3000) = 290200 + 90 = 290 checks$5000 is invested at 4%, $3000 is invested at 3%.- .--...Show more