Question

19. A group of students takes their math teacher out for lunch. If five have hamburgers and three have hot dogs, the bill will be $24.25 If three have hamburgers and five have hot dogs, the bill will be $21.75. What is the price of a hamburger? Show all work ​

Answer 1

Answer:

$3.50 = price of hamburger

Skills needed: Algebraic Thinking, Systems of Equations

Step-by-step explanation:

1) Let's create equations to represent this problem.

• Defining variables: $b=hamburger price, h=hotdogprice$d
• Creating equations:

First one is that 5 hamburgers + 3 hot dogs is 24.25

----> $5b+3h=24.25$

Second one is that 3 hamburgers + 5 hot dogs is 21.75

----> $3b+5h=21.75$

Our two equations are:

- $5b+3h=24.25 \\ 3b+5h=21.75$

2) Solving this out:

Let's use elimination. Since we are trying to solve for $b$, we need to eliminate $h$. We need to find the LCM of 3 and 5, which is 15 since both are prime numbers (if both are prime, multiply together to get the LCM)

a. $5(5b+3h)=5(24.25) \\ 3(3b+5h)=5(21.75)$

For the 1st equation: $5*5b=25b, 5*3h=15h, 24.25*5=121.25$

New first equation: $25b+15h=121.25$

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For the 2nd equation: $3*3b=9b, 3*5h=15h, 21.75*3=65.25$

New second equation: $9b+15h=65.25$

b. Now we eliminate h by subtraction:

$\text{ }+ 25b+15h=121.25 \\ - (+9b + 15h) = 65.25$

We combine them both while subtracting the second from the first.

On the right side: $121.25-65.25 = 56$

On the left side: $25b-9b=16b$, $15h-15h=0 \\\text{0 means nothing so it's removed}$

c. We solve for $b$

$16b=56$

$b=3.5$

A hamburger is $3.50 based on the above.