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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineMrs. Chapman sometimes makes her melon salad for special events. When she made it a couple months ago, she used 11 kilogram of honeydew melon and 22 kilograms of watermelon, which cost her $9\$9. Today, she used 22 kilograms of honeydew melon and 11 kilogram of watermelon, spending a total of $9\$9 on the melons. Assuming that the prices of the melons haven't changed, how much does a kilogram of each type of melon cost?\newlineHoneydew melon costs $\$_____ per kilogram and watermelon costs $\$_____ per kilogram.

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineMrs. Chapman sometimes makes her melon salad for special events. When she made it a couple months ago, she used 11 kilogram of honeydew melon and 22 kilograms of watermelon, which cost her $9\$9. Today, she used 22 kilograms of honeydew melon and 11 kilogram of watermelon, spending a total of $9\$9 on the melons. Assuming that the prices of the melons haven't changed, how much does a kilogram of each type of melon cost?\newlineHoneydew melon costs $\$_____ per kilogram and watermelon costs $\$_____ per kilogram.
  1. Write Equations: Let's denote the cost of 11 kilogram of honeydew melon as HH dollars and the cost of 11 kilogram of watermelon as WW dollars. We can write two equations based on the given information.\newlineFirst event: 1H+2W=($)91H + 2W = (\$)9\newlineSecond event: 2H+1W=($)92H + 1W = (\$)9
  2. Eliminate Variable: To solve the system of equations using elimination, we need to eliminate one of the variables. We can do this by multiplying the first equation by 22 and the second equation by 11, so that when we subtract the second equation from the first, the WW terms will cancel out.\newlineFirst equation multiplied by 22: (1H+2W)×2=($9)×2(1H + 2W) \times 2 = (\$9) \times 2\newlineThis gives us: 2H+4W=($18)2H + 4W = (\$18)\newlineSecond equation multiplied by 11: (2H+1W)×1=($9)×1(2H + 1W) \times 1 = (\$9) \times 1\newlineThis gives us: 2H+1W=($9)2H + 1W = (\$9)
  3. Subtract Equations: Now we subtract the second equation from the first to eliminate WW:(2H+4W)−(2H+1W)=($18)−($9)(2H + 4W) - (2H + 1W) = (\$18) - (\$9)This simplifies to: 3W=($9)3W = (\$9)
  4. Solve for W: Divide both sides of the equation by 33 to solve for W:\newline3W3=$93\frac{3W}{3} = \frac{\$9}{3}\newlineThis gives us: W=$3W = \$3
  5. Substitute and Solve: Now that we know the cost of watermelon per kilogram, we can substitute W=$3W = \$3 into one of the original equations to find HH. We'll use the first equation:\newline1H+2W=$91H + 2W = \$9\newlineSubstitute W=$3W = \$3:\newline1H+2($3)=$91H + 2(\$3) = \$9
  6. Substitute and Solve: Now that we know the cost of watermelon per kilogram, we can substitute W=$3W = \$3 into one of the original equations to find HH. We'll use the first equation:\newline1H+2W=$91H + 2W = \$9\newlineSubstitute W=$3W = \$3:\newline1H+2($3)=$91H + 2(\$3) = \$9Solve for HH:\newline1H+$6=$91H + \$6 = \$9\newlineSubtract \$\(6\) from both sides:\(\newline\)\[1H = \$9 - \$6\]\(\newline\)This gives us: \(H = \$3\)

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