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How many gallons each of 30% alcohol and 10% alcohol should be mixed to obtain 20 gal of 15% alcohol? Gallons of Solution Percent (as a decimal) Gallons of Pure Alcohol x 30%=0.3 y 10%=0.1 20 15%= How many gallons of 30% alcohol should be in the mixture? ◻ gal How many gallons of 10% alcohol should be in the mixture? ◻ gal

How many gallons each of 30% 30 \% alcohol and 10% 10 \% alcohol should be mixed to obtain 2020 gal of 15% 15 \% alcohol?\newline\begin{tabular}{|c|c|c|}\newline\hline \begin{tabular}{c} \newlineGallons of \\\newlineSolution\newline\end{tabular} & \begin{tabular}{c} \newlinePercent \\\newline(as a decimal)\newline\end{tabular} & \begin{tabular}{c} \newlineGallons of \\\newlinePure \\\newlineAlcohol\newline\end{tabular} \\\newline\hlinex x & 30%=0.3 30 \%=0.3 & \\\newline\hliney y & 10%=0.1 10 \%=0.1 & \\\newline\hline 2020 & 15%= 15 \%= & \\\newline\hline\newline\end{tabular}\newlineHow many gallons of 30% 30 \% alcohol should be in the mixture? □ \square gal\newlineHow many gallons of 10% 10 \% alcohol should be in the mixture? □ \square gal

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Q. How many gallons each of 30% 30 \% alcohol and 10% 10 \% alcohol should be mixed to obtain 2020 gal of 15% 15 \% alcohol?\newline\begin{tabular}{|c|c|c|}\newline\hline \begin{tabular}{c} \newlineGallons of \\\newlineSolution\newline\end{tabular} & \begin{tabular}{c} \newlinePercent \\\newline(as a decimal)\newline\end{tabular} & \begin{tabular}{c} \newlineGallons of \\\newlinePure \\\newlineAlcohol\newline\end{tabular} \\\newline\hlinex x & 30%=0.3 30 \%=0.3 & \\\newline\hliney y & 10%=0.1 10 \%=0.1 & \\\newline\hline 2020 & 15%= 15 \%= & \\\newline\hline\newline\end{tabular}\newlineHow many gallons of 30% 30 \% alcohol should be in the mixture? □ \square gal\newlineHow many gallons of 10% 10 \% alcohol should be in the mixture? □ \square gal
  1. Define variables: Let xx be the gallons of 30%30\% alcohol, and yy be the gallons of 10%10\% alcohol.
  2. Total volume equation: The total volume of the mixture should be 2020 gallons, so x+y=20x + y = 20.
  3. Total alcohol amount: The total amount of pure alcohol in the mixture should be 15%15\% of 2020 gallons, so 0.15×20=30.15 \times 20 = 3 gallons.
  4. Alcohol content equations: The amount of pure alcohol in the 30%30\% solution is 0.3x0.3x and in the 10%10\% solution is 0.1y0.1y.
  5. Combine equations: The sum of pure alcohol from both solutions should equal the total pure alcohol in the mixture: 0.3x+0.1y=30.3x + 0.1y = 3.
  6. Substitute and simplify: Now we have two equations: x+y=20x + y = 20 and 0.3x+0.1y=30.3x + 0.1y = 3.
  7. Solve for x: Solve the first equation for y: y=20−xy = 20 - x.
  8. Solve for x: Solve the first equation for yy: y=20−xy = 20 - x.Substitute yy in the second equation: 0.3x+0.1(20−x)=30.3x + 0.1(20 - x) = 3.
  9. Solve for x: Solve the first equation for y: y=20−xy = 20 - x. Substitute yy in the second equation: 0.3x+0.1(20−x)=30.3x + 0.1(20 - x) = 3. Distribute and combine like terms: 0.3x+2−0.1x=30.3x + 2 - 0.1x = 3.
  10. Solve for x: Solve the first equation for y: y=20−xy = 20 - x. Substitute yy in the second equation: 0.3x+0.1(20−x)=30.3x + 0.1(20 - x) = 3. Distribute and combine like terms: 0.3x+2−0.1x=30.3x + 2 - 0.1x = 3. Combine xx terms: 0.2x+2=30.2x + 2 = 3.
  11. Solve for x: Solve the first equation for y: y=20−xy = 20 - x.Substitute y in the second equation: 0.3x+0.1(20−x)=30.3x + 0.1(20 - x) = 3.Distribute and combine like terms: 0.3x+2−0.1x=30.3x + 2 - 0.1x = 3.Combine x terms: 0.2x+2=30.2x + 2 = 3.Subtract 22 from both sides: 0.2x=10.2x = 1.
  12. Solve for x: Solve the first equation for y: y=20−xy = 20 - x. Substitute yy in the second equation: 0.3x+0.1(20−x)=30.3x + 0.1(20 - x) = 3. Distribute and combine like terms: 0.3x+2−0.1x=30.3x + 2 - 0.1x = 3. Combine xx terms: 0.2x+2=30.2x + 2 = 3. Subtract 22 from both sides: 0.2x=10.2x = 1. Divide by 0.20.2 to solve for xx: yy00.

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