document.write( "Question 553609: one solution is 20% acid and another is 70% acid. How many liters of each solution should you use to make 4 liters of a 55% solution \n" ); document.write( "
Algebra.Com's Answer #360961 by mananth(16946)![]() ![]() You can put this solution on YOUR website! percent ----------------quantity \n" ); document.write( "Acid type I 20 ----------------x liters \n" ); document.write( "Acid type II 70 ----------------4-x liters \n" ); document.write( "Mixture 55.00% ---------------- 4 \n" ); document.write( " \n" ); document.write( "20x+70(4-x)=55*4 \n" ); document.write( " \n" ); document.write( "20x+280-70x=220 \n" ); document.write( "20x-70x=220-280 \n" ); document.write( "-50x=-60 \n" ); document.write( "/-50 \n" ); document.write( "x=1.2 liters 20.00% Acid type I \n" ); document.write( "2.8 liters 70.00% Acid type II \n" ); document.write( " \n" ); document.write( " |