document.write( "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?\r
\n" ); document.write( "\n" ); document.write( "Can you show me in steps, so I will know how to solve? I would appreciate it. I need to know how to solve.\r
\n" ); document.write( "\n" ); document.write( "Thank you.
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Algebra.Com's Answer #190205 by richwmiller(17219)\"\" \"About 
You can put this solution on YOUR website!
x+y=10,000
\n" ); document.write( "y=10000-x
\n" ); document.write( ".06*x+.075*y=721.50
\n" ); document.write( "substitute 10000-x for y\r
\n" ); document.write( "\n" ); document.write( ".06*x+.075*(10000-x)=721.50
\n" ); document.write( "solve for x then plug into y=10000-x for y\r
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