Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ Mrs. Chapman sometimes makes her melon salad for special events. When she made it a couple months ago, she used $1$ kilogram of honeydew melon and $2$ kilograms of watermelon, which cost her $\$9$ . Today, she used $2$ kilograms of honeydew melon and $1$ kilogram of watermelon, spending a total of $\$9$ on the melons. Assuming that the prices of the melons haven't changed, how much does a kilogram of each type of melon cost? $\newline$ Honeydew melon costs $\$$ _ per kilogram and watermelon costs $\$$ _ per kilogram.

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Grade 8

Solve a system of equations using elimination: word problems

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.

\newline

Mrs. Chapman sometimes makes her melon salad for special events. When she made it a couple months ago, she used

1

kilogram of honeydew melon and

2

kilograms of watermelon, which cost her

\$9

. Today, she used

2

kilograms of honeydew melon and

1

kilogram of watermelon, spending a total of

\$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?

\newline

Honeydew melon costs

\$

_____ per kilogram and watermelon costs

\$

_____ per kilogram.

Write Equations: Let's denote the cost of $1$ kilogram of honeydew melon as $H$ dollars and the cost of $1$ kilogram of watermelon as $W$ dollars. We can write two equations based on the given information. $\newline$ First event: $1H + 2W = (\$)9$ $\newline$ Second event: $2H + 1W = (\$)9$
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 $2$ and the second equation by $1$ , so that when we subtract the second equation from the first, the $W$ terms will cancel out. $\newline$ First equation multiplied by $2$ : $(1H + 2W) \times 2 = (\$9) \times 2$ $\newline$ This gives us: $2H + 4W = (\$18)$ $\newline$ Second equation multiplied by $1$ : $(2H + 1W) \times 1 = (\$9) \times 1$ $\newline$ This gives us: $2H + 1W = (\$9)$
Subtract Equations: Now we subtract the second equation from the first to eliminate $W$ : $(2H + 4W) - (2H + 1W) = (\$18) - (\$9)$ This simplifies to: $3W = (\$9)$
Solve for W: Divide both sides of the equation by $3$ to solve for W: $\newline$ $\frac{3W}{3} = \frac{\$9}{3}$ $\newline$ This gives us: $W = \$3$
Substitute and Solve: Now that we know the cost of watermelon per kilogram, we can substitute $W = \$3$ into one of the original equations to find $H$ . We'll use the first equation: $\newline$ $1H + 2W = \$9$ $\newline$ Substitute $W = \$3$ : $\newline$ $1H + 2(\$3) = \$9$
Substitute and Solve: Now that we know the cost of watermelon per kilogram, we can substitute $W = \$3$ into one of the original equations to find $H$ . We'll use the first equation: $\newline$ $1H + 2W = \$9$ $\newline$ Substitute $W = \$3$ : $\newline$ $1H + 2(\$3) = \$9$ Solve for $H$ : $\newline$ $1H + \$6 = \$9$ $\newline$ Subtract \$$6$ from both sides:$\newline$\[1H = \$9 - \$6\]$\newline$This gives us: $H = \$3$