assignment 9.2: systems of linear equations
score: 8/9 answered: 8/9
question 9
a chemist has three different acid solutions. the first acid solution contains 20% acid, the second contains 35% acid, and the third contains 75% acid. they want to use all three solutions to obtain a mixture of 126 liters of a solution as the 35% solution, using 3 times as much of the 75% solution as the 40% acid solution. how many liters of each solution should be used?
the chemist should use liters of 20% solution, liters of 35% solution, liters of 75% solution.
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Discipline: Mathematics
Sub - field: Algebra
Answer: Let $x$ be the liters of the 20% acid solution, $y$ be the liters of the 35% acid solution, and $z$ be the liters of the 75% acid solution. We know $z = 3y$ and $x + y+z=126$, and $0.2x + 0.35y+0.75z=0.4\times126$. Substitute $z = 3y$ into the other two equations: $x + y + 3y=126$ (so $x + 4y=126$ or $x=126 - 4y$) and $0.2x+0.35y + 0.75\times3y=50.4$. Substitute $x = 126-4y$ into $0.2x+0.35y + 2.25y=50.4$: $0.2(126 - 4y)+2.6y=50.4$, $25.2-0.8y+2.6y=50.4$, $1.8y=25.2$, $y = 14$. Then $z = 3y=42$ and $x=126-(14 + 42)=70$. So the chemist should use 70 liters of 20% solution, 14 liters of 35% solution, and 42 liters of 75% solution.

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