Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ At a community barbecue, Mrs. Callahan and Mr. Maynard are buying dinner for their families. Mrs. Callahan purchases $3$ hot dog meals and $2$ hamburger meals, paying a total of $32. Mr. Maynard buys $2$ hot dog meals and $1$ hamburger meal, spending $19$ $ in all. How much do the meals cost? $\newline$ Hot dog meals cost $\$\_\_\_\_$ each, and hamburger meals cost $\$\_\_\_\_$ each.

Full solution

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

\newline

At a community barbecue, Mrs. Callahan and Mr. Maynard are buying dinner for their families. Mrs. Callahan purchases

3

hot dog meals and

2

hamburger meals, paying a total of \(32. Mr. Maynard buys

2

hot dog meals and

1

hamburger meal, spending

19

\) in all. How much do the meals cost?

\newline

Hot dog meals cost

\$\_\_\_\_

each, and hamburger meals cost

\$\_\_\_\_

each.

Equation $1$ : Mrs. Callahan's purchase: Let's denote the price of each hot dog meal as $h$ and the price of each hamburger meal as $d$ . We can write two equations based on the information given. The first equation comes from Mrs. Callahan's purchase: $3$ hot dog meals and $2$ hamburger meals for $\$32$ , which gives us the equation $3h + 2d = 32$ .
Equation $2$ : Mr. Maynard's purchase: The second equation comes from Mr. Maynard's purchase: $2$ hot dog meals and $1$ hamburger meal for $\$19$ , which gives us the equation $2h + d = 19$ .
Eliminating variable ' extit{d}': We have a system of two equations now. To use elimination, we need to make the coefficients of one of the variables the same in both equations. We decide to eliminate extit{d}. To do this, we multiply the second equation by $2$ to match the coefficient of extit{d} in the first equation, resulting in $4h + 2d = 38$ .
Solving for 'h': Now we subtract the first equation from the new second equation to eliminate $d$ . This gives us $(4h + 2d) - (3h + 2d) = 38 - 32$ , which simplifies to $h = 6$ .
Substituting 'h' into Equation $1$ : With the value of $h$ found, we substitute $h = 6$ into the first equation to solve for $d$ . Substituting into $3h + 2d = 32$ gives us $3(6) + 2d = 32$ , which simplifies to $18 + 2d = 32$ .
Solving for 'd': Subtracting $18$ from both sides of the equation $18 + 2d = 32$ gives us $2d = 14$ . Dividing both sides by $2$ gives us $d = 7$ .