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Unformatted text preview: 100 % x L 100/1 L 100 x 30 % 6 L 30/1 L 180 50 % x + 6 L 50/1 L 50 ( x + 6) Since the pure acid added to the 30 % acid must equal the 50 % acid, the governing equation for the problem is: 100x + 180 = 50(x + 6) 1 IV. Solution of the Mathematical Problem Using the distributive property on the above equation gives 100x + 180 = 300 + 50x Subtract 50x from both sides of the equation 50x + 180 = 300 Subtract 180 from both sides of the equation 50x = 120 Divide both sides of the equation by 50 to solve for x x = 2.4 For the amount of 50 % acid, adding 6 to 2.4 gives 8.4. V. Interpretation of the Mathematical Results In the problem, x was the amount of pure acid needed to add to 6 liters of 30 % acid to create 50 % acid. Therefore, adding 2.4 liters of 100 % acid to 6 liters of 30 % acid will create 8.4 liters of 50 % acid. 2...
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This note was uploaded on 05/16/2008 for the course MATH H1513 taught by Professor Walker during the Fall '08 term at East Central.
 Fall '08
 Walker
 Algebra, Distance Formula

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