document.write( "Question 910464: How many gallons of 70% antifreeze should be mixed with 10 gallons of 30% antifreeze to obtain a 45% antifreeze mixture?\r
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document.write( "Here's what I have
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document.write( "(0.30)(10gal)+(0.70)(x)=(0.45)(10+x)
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document.write( "After I solve this I get 6 gallons but when I check the solution I don't get 45% so this cannot be the right way to solve this problem. Please help. \n" );
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Algebra.Com's Answer #552545 by richwmiller(17219)![]() ![]() You can put this solution on YOUR website! We have 10 gal of 30% antifreeze \n" ); document.write( "We want to add some 70% antifreeze to make 45% antifreeze \n" ); document.write( "0.7*x+0.3*10=0.45(10+x) \n" ); document.write( "0.7*x+3=4.5+0.45x \n" ); document.write( "0.7x-0.45x=4.5-3 \n" ); document.write( "0.25x=1.5 \n" ); document.write( "x=6 gal at 70% (4.2) is added to 10 gal at 30%(3) to make 45% of 16 gal(7.2) \n" ); document.write( "check \n" ); document.write( "Remember it is 45% of 16 gallons that is .45*16 which is 7.2 \n" ); document.write( "0.7*6+0.3*10=0.45(10+6) \n" ); document.write( "4.2+3=0.45(16) \n" ); document.write( "7.2=7.2 \n" ); document.write( "ok \n" ); document.write( " \n" ); document.write( " |