document.write( "Question 160573: Stella's catering is planning a wedding reception. The bride and groom would like to serve a nut mixture containing 25% peanuts. Stella has available mixtures that are either 40% or 10% peanuts. How much of each type should be mixed to get a 10-lb mixture that is 25% peanuts? \n" ); document.write( "

Algebra.Com's Answer #118404 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x=amount of 40% peanuts needed
\n" ); document.write( "Then 10-x=amount of 10% peanuts needed\r
\n" ); document.write( "\n" ); document.write( "Amount of pure peanuts in the 40% mixture (0.40x) plus amount of pure peanuts in the 10% mixture ((0.10(10-x)) has to equal amount of pure peanuts in final mixture(0.25*10). So, our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.40x+0.10(10-x)=0.25*10 get rid of parens\r
\n" ); document.write( "\n" ); document.write( "0.40x+1-0.10x=2.5 subtract 1 from each side
\n" ); document.write( "0.40x-0.10x+1-1=2.5-1 collect like terms
\n" ); document.write( "0.30x=1.5 divide each side by 0.30
\n" ); document.write( "x=5 lb----------------------------------amount of 40% peanuts needed
\n" ); document.write( "10-x=10-5=5 lb---------------amount of 10% peanuts needed\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "0.40*5+0.10*5=0.25*10
\n" ); document.write( "2+0.5=2.5
\n" ); document.write( "2.5=2.5\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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