document.write( "Question 136750This question is from textbook Algebra 1
\n" ); document.write( ": A solution containing 30% insecticide is to be mixed with a solution containing 50% insecticide to make 200 L of a solutions containing 42% insecticide. How much of each solution should be used? \n" ); document.write( "
Algebra.Com's Answer #100108 by ptaylor(2198)\"\" \"About 
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Let x=amount of 30% insecticide needed
\n" ); document.write( "Then 200-x=amount of 50% insecticide needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure insecticide in the 30% solution(0.30x) plus the amount of pure insecticide in the 50% solution(0.50(200-x)) has to equal the amount of pure insecticide in the final mixture (0.42*200). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.30x+0.50(200-x)=0.42*200 get rid of parens\r
\n" ); document.write( "\n" ); document.write( "0.30x+100-0.50x=84 subtract 100 from each side\r
\n" ); document.write( "\n" ); document.write( "0.30x+100-100-0.50x=84-100 collect like terms\r
\n" ); document.write( "\n" ); document.write( "-0.20x=-16 divide both sides by -0.20\r
\n" ); document.write( "\n" ); document.write( "x=80 L----------------------amount of 30% insecticide needed\r
\n" ); document.write( "\n" ); document.write( "200-x=200-80=120 L---------------------amount of 50% insecticide needed\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "0.30*80+0.50*120=0.42*200\r
\n" ); document.write( "\n" ); document.write( "24+60=84
\n" ); document.write( "84=84\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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