Algebra.Com's Answer #656437 by MathTherapy(10552)  You can put this solution on YOUR website! \n" );
document.write( "You have 23 coins totaling $2.75, how many quarters, nickles, and dimes do you have? \n" );
document.write( "Let number of quarters, dimes, and nickels be Q, D, and N, respectively \n" );
document.write( "Then we get: Q + D + N = 23 -------- eq (i) \n" );
document.write( "Also, .25Q + .1D + .05N = 2.75 ----- eq (ii) \n" );
document.write( "Since there are a lot of coins worth only $2.75, there MUST be a small number of quarters. Therefore, we start with 1 quarter, but 1, 2, and 3 \n" );
document.write( "quarters DO NOT result in meaningful numbers of nickels, or dimes. However, 4 quarters do. \n" );
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document.write( "Let number of quarters be 4 \n" );
document.write( "Then we get: 4 + D + N = 23____D + N = 19 -------- eq (iii) \n" );
document.write( ".25(4) + .1D + .05N = 2.75_____1 + .1D + .05N = 2.75____.1D + .05N = 1.75 ------- eq (iv) \n" );
document.write( "- .1D - .1N = - 1.9 ----- Multiplying eq (iii) by - .1 ------- eq (v) \n" );
document.write( "- .05N = - .15 ---------- Adding eqs (v) & (iv) \n" );
document.write( "N, or number of nickels = , or 3\r \n" );
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document.write( "4 + D + 3 = 23 -------- Substituting 4 for Q and 3 for N in eq (i) \n" );
document.write( "D + 7 = 23 \n" );
document.write( "D, or number of dimes = 23 – 7, or 16\r \n" );
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document.write( "This gives us: \n" );
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