document.write( "Question 1085556: How many gallons of 80% antifreeze solution must be mixed with 100 gallons of 15% antifreeze to get a mixture that is 70% antifreeze?\r
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Algebra.Com's Answer #699637 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
let x = the number of gallons of 80% antifreeze.\r
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\n" ); document.write( "\n" ); document.write( "your formula becomes:\r
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\n" ); document.write( "\n" ); document.write( ".8 * x + .15 * 100 = .7 * (x + 100)\r
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\n" ); document.write( "\n" ); document.write( "simplify this formula to get .8 * x + 15 = .7 * x + 70\r
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\n" ); document.write( "\n" ); document.write( "subtract .7 * x from both sides of the equation and subtract 15 from both sides of the equation to get .8 * x - .7 * x = 70 - 15\r
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\n" ); document.write( "\n" ); document.write( "simplify to get .1 * x = 55.\r
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\n" ); document.write( "\n" ); document.write( "solve for x to get x = 55/.1 = 550 gallons.\r
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\n" ); document.write( "\n" ); document.write( "you would need 550 gallons of 80% antifreeze to mix with 100 gallons of 15% antifreeze to get 650 gallons of 70% antifreeze.\r
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\n" ); document.write( "\n" ); document.write( ".8 * 550 = 440 gallons of pure antifreeze
\n" ); document.write( ".15 * 100 = 15 gallons of pure antifreeze
\n" ); document.write( "total = 455 gallons of pure antifreeze\r
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\n" ); document.write( "\n" ); document.write( ".7 * 650 = 455 gallons of pure antifreeze\r
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\n" ); document.write( "\n" ); document.write( "solution looks good.\r
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\n" ); document.write( "\n" ); document.write( "you would need 550 gallons of 80% antifreeze solution.\r
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