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? Can you show me in

Algebra ->  Linear-equations -> 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? Can you show me in      Log On


   



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) About Me  (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