document.write( "Question 15069: Mitch invested $7500 for 1 year, part at 12% annual intrest and the rest at 10% annual intreast. His total intrest for one year was 890. How much money did he invest at 12%. \n" ); document.write( "
Algebra.Com's Answer #7504 by bam878s(77)\"\" \"About 
You can put this solution on YOUR website!
Let i = interest rate p = amount at end of year, s = amount at beginning of year, and t = time in years.
\n" ); document.write( "Now depending on whether you are using simple or compound interest the equations are
\n" ); document.write( "p = x\"%281%2Bi%29%5Et\" for compound interest and
\n" ); document.write( "p = x(1+it) for simple interest
\n" ); document.write( "Note these two formulas are the same for t = 1 year so it doesn't matter which one we use. I will use the simple interest formula since I don't have to deal with solving exponents.\r
\n" ); document.write( "\n" ); document.write( "Total interest = 890
\n" ); document.write( "x = part invested at 12%
\n" ); document.write( "y = part invested at 10%
\n" ); document.write( "Amount in account at end of year is 7500 + 890 = 8390.
\n" ); document.write( "So 8390 = 1.12x + 1.10y. (1)
\n" ); document.write( "and 7500 = x + y. (2)
\n" ); document.write( "multiply (2) by 1.12 TO GET
\n" ); document.write( "8400 = 1.12X + 1.10Y
\n" ); document.write( "Now subtract (1) from this to get
\n" ); document.write( "(8400-8390) = (1.12-1.12)x + (1.12-1.10)y
\n" ); document.write( "10 = .02y
\n" ); document.write( "now divide both sides by .02 to get
\n" ); document.write( "y = 500 = amount invested at 10%
\n" ); document.write( "Now plug this back into either (1) or (2) to solve for x
\n" ); document.write( "x + 500 = 7500
\n" ); document.write( "x = 7000 = amount invested at 12%
\n" ); document.write( "This would make since since you more than likely invest more money at the higher interest rate to yield a higher return.
\n" ); document.write( " \n" ); document.write( "
\n" );