Question

Three sodas and one hamburger cost $4.50. Two sodas and 2 hamburgers cost $4.00. One soda costs:

Answer 1

$1.25

Explanation:

1)We can set two equatons for each statement.

Three sodas and one hamburger cost $4.50:

3s + h = 4.5

Two sodas and 2 hamburgers cost $4.00:

2s + 2h = 4

2) Solve using simultaneous equations. To find out the cost of one soda we need to make the number of hamburgers the same in both equations to eliminate them.

3s + h = 4.5

2s + 2h = 4

To do this multiply the first equation across by 2 because in the second equation there is 2h, which also makes the first equation 2h. This gives us:

6s + 2h = 9

2s + 2h = 4

3) subtract across to eliminate the h.

6s - 2s = 4s, 2h - 2h = 0h, 9 - 4 = 5, so we now have:

4s = 5

4) divide by 4 to solve for s:

s = 5/4 = $1.25

Answer 2

1) We can solve this problem using a system of equations. So let's cal h for hamburger and s for sodas.

2) So let's set the system:

`\begin{gathered} 3s+h=4.5 \\ 2s+2h=4 \end{gathered}`

Hence, we can solve them by the Elimination Method.

Let's multiply the first equation by -2 to eliminate the h terms

Since the question is about the soda's price, we can stop it here.