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Marine Biology. A marine biologist wants to set up an aquarium containing $3 \%$ salt water. He has two tanks on hand that contain $6 \%$ and $2 \%$ salt water. How much water from each tank must he use to fill a 32 -gallon aquarium with a $3 \%$ saltwater mixture?
8 gal $6 \%$ salt water, 24 gal $2 \%$ salt water
Algebra
Chapter 4
Systems of Linear Equations and Inequalities
Section 4
Problem Solving Using Systems of Equations
Equations and Inequalities
Systems of Equations and Inequalities
Campbell University
McMaster University
University of Michigan - Ann Arbor
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Okay, so even take. It's going to be a 32 gallon aquarium with 3% salt water. So that's the goal. Go equals 32 gallons and 3% and what we have. We have two different options. A 6% tank and a 2% think so. Maybe that from 6% and 2%. So let's just say X is the amount of 6%. And why's the amount of 2% water we're gonna use? So we know two things. We know that there is a total of 32 gallons going into the tank, and we also know how much salt we want. The amount of salt that we want total is 3% of 32. So you just take 320.3 and multiply that by 32 and that gives you 96.96. Make that six a little bit of Queen 96. So that's my other piece of information. 6% of acts in 2% of why makes 0.96 looks to me like a good elimination problem gonna multiply the top equation. My 0.2 That's the coefficient of why cause then I could write it down here below 0.2 x plus 0.2 Why and then 0.2 times 32 is 320.64. Now we're ready to eliminate. Subtracting the two equations and the wise cancel out 0.6 minus 0.0 to 6.4 X 0.96 minus 0.64. This 0.32 did I need my 0.4 gives you the value for X X equals eight. You know that there's 32 gallons total. So why has to equal the rest? 32 minus eight is 24 just as a little nice.
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