document.write( "Question 1194955: Mr. and Mrs. Jackson invested some money at 6% simple interest and some money at 10% simple interest. In the second investment they put $1000.00 more than they put in the first. If the income from both investments for 1 year was $340.00, then how much did they invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #827256 by Theo(13342)![]() ![]() You can put this solution on YOUR website! x = amount invested at 6%. \n" ); document.write( "x + 1000 = amount invested at 10%. \n" ); document.write( ".06x + .1 * (x + 1000) = 340 \n" ); document.write( "simplify to get: \n" ); document.write( ".06x + .1x + 100 = 340 \n" ); document.write( "subtract 100 from both sides of the equation and combine like terms to get: \n" ); document.write( ".16x = 240 \n" ); document.write( "solve for x to get: \n" ); document.write( "x = 240/.16 = 24000/16 = 3000 / 2 = 1500. \n" ); document.write( ".06 * 1500 + .1 * (1500 + 1000) = .06 * 1500 + .1 * 2500 = 90 + 250 = 340. \n" ); document.write( "this confirms the value of x = 1500 is correct. \n" ); document.write( "solution is they invested 1500 at 6% and 2500 at 10%. \n" ); document.write( " \n" ); document.write( " |