document.write( "Question 679506: Solving Linear Systems of equations
\n" ); document.write( "The problem: one is 5% acid and one is 6.5% acid . You want to make 200ml of vinegar,with 6% acid. How many ml of each vinegar do you need to mix together?\r
\n" ); document.write( "\n" ); document.write( "My question: With percents I get confused, and I am not sure which variables you would use to set this up into two equations? \n" ); document.write( "

Algebra.Com's Answer #421967 by mananth(16946) $\"\"$ $\"About$

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Let quantity of 5% acid required be x
\n" ); document.write( "Let quantity of 6.5% acid required be y\r
\n" ); document.write( "\n" ); document.write( "Mixture = 6%\r
\n" ); document.write( "\n" ); document.write( "Equate percentages\r
\n" ); document.write( "\n" ); document.write( "5%x+6.5%y=6%*200
\n" ); document.write( "multiply by 100
\n" ); document.write( "5x+6.5y=1200................(1)\r
\n" ); document.write( "\n" ); document.write( "Quantity
\n" ); document.write( "x+y=200.....................(2)\r
\n" ); document.write( "\n" ); document.write( "multiply equation (2) by -5
\n" ); document.write( "we get -5x-5y=-1000
\n" ); document.write( "Add this equation to (1)\r
\n" ); document.write( "\n" ); document.write( "we get
\n" ); document.write( "1.5y=200
\n" ); document.write( "y= 200/1.5
\n" ); document.write( "y=133.3 ml
\n" ); document.write( "x= balance amount= 66.67ml \n" ); document.write( "

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