document.write( "Question 168604: Kathy has 5 liters of a 32% acid solution and she4 also has a large amount of a 26% acid solution. How many liters of the 26% solution can Kathy mix with the 5 liters of 32% solution in order to produce a 50% acid solution?\r
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document.write( "the answer isn't possible in the end but i need to know the values of x and y and write out a system of equations from the information given and then solve hte system of equations using substitution or elimination \n" );
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Algebra.Com's Answer #124261 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Kathy has 5 liters of a 32% acid solution and she also has a large amount of a \n" ); document.write( " 26% acid solution. How many liters of the 26% solution can Kathy mix with \n" ); document.write( "the 5 liters of 32% solution in order to produce a 50% acid solution? \n" ); document.write( ": \n" ); document.write( "I think you would just use a typical mixture equation, if it's impossible, it \n" ); document.write( " will let you know, by giving a negative solution. \n" ); document.write( ": \n" ); document.write( "Let x = amt of 26% solution (why would you need y?) \n" ); document.write( ": \n" ); document.write( ".32(5) + .26x = .50(x+5) \n" ); document.write( ": \n" ); document.write( "1.6 + .26x = .50x = 2.5 \n" ); document.write( ": \n" ); document.write( ".26x - .50x = 2.5 - 1.6 \n" ); document.write( "-.24x = .9 \n" ); document.write( "x = \n" ); document.write( "x = -3.75; clearly, not of the real world \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Substitution will show the equation is satisfied, but negative substances don't exist \n" ); document.write( " |