SOLUTION: Part of $10,000 was invested at 6% interest and the rest at 7.5%. If the annual income from these investments was $712.50, how much was invested at each rate?
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-> SOLUTION: Part of $10,000 was invested at 6% interest and the rest at 7.5%. If the annual income from these investments was $712.50, how much was invested at each rate?
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Question 258368: Part of $10,000 was invested at 6% interest and the rest at 7.5%. If the annual income from these investments was $712.50, how much was invested at each rate?
Can you show me in steps, so I will know how to solve? I would appreciate it. I need to know how to solve.
Thank you. Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! x+y=10,000
y=10000-x
.06*x+.075*y=721.50
substitute 10000-x for y
.06*x+.075*(10000-x)=721.50
solve for x then plug into y=10000-x for y