document.write( "Question 147450: If grade A coffee costs .95 cents a pound and grade b coffee costs .75 cents a pound, how many pounds of each grade is needed if the total cost is $50.50, and the amount of grade b that is needed is twice the amount needed for grad A plus 5 pounds? \n" ); document.write( "
Algebra.Com's Answer #107837 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
If grade A coffee costs .95 cents a pound and grade b coffee costs .75 cents a pound, how many pounds of each grade is needed if the total cost is $50.50, and the amount of grade b that is needed is twice the amount needed for grad A plus 5 pounds?
\n" ); document.write( ":
\n" ); document.write( "Let x = amt of grade a coffee required
\n" ); document.write( ":
\n" ); document.write( "It says,\"the amount of grade b that is needed is twice the amount needed for grad A plus 5 pounds\", therefore we can say:
\n" ); document.write( "2x + 5 = amt of grade b coffee required
\n" ); document.write( ":
\n" ); document.write( "total cost equation;
\n" ); document.write( ".95x + .75(2x+5) = 50.50
\n" ); document.write( ":
\n" ); document.write( ".95x + 1.5x + 3.75 = 50.50
\n" ); document.write( ":
\n" ); document.write( "2.45x = 50.50 - 3.75
\n" ); document.write( ":
\n" ); document.write( "2.45x = 46.75
\n" ); document.write( "x = \"46.75%2F2.45\"
\n" ); document.write( "x = 19.08 lb of grade a
\n" ); document.write( "and
\n" ); document.write( "2(19.08) + 5 = 43.16.lb of grade b
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check solution on a calc:
\n" ); document.write( ".95(19.08) + .75(43.16) =
\n" ); document.write( " 18.126 + 32.37 = 50.496 ~ 50.50
\n" ); document.write( ":
\n" ); document.write( "Did this make sense to you? Any questions?
\n" ); document.write( " \n" ); document.write( "
\n" );