Question

8. Three orders are placed at a pizza shop. Two small pizzas, a liter of pop and a salad cost $28; onesmall pizza, a liter of pop, and three salads cost $30; and three small pizzas, a liter of pop, and two saladscost $44. How much does each item cost?

Answer 1

Let

small pizza cost "p"

liter of pop cost "x"

salad cost "y"

From the three statements, we can write 3 equations:

1. Two small pizzas, a liter of pop and a salad cost $28:

$2p+x+y=28$

2. one small pizza, a liter of pop, and three salads cost $30:

$p+x+3y=30$

3. three small pizzas, a liter of pop, and two salads cost $44:

$3p+x+2y=44$

We need to solve these 3 simultaneous equations in order to find the value of "p", "x", and "y".

Multiplying 2nd equation by (-1) and adding it to 1st equation, we get:

$\begin{gathered} 2p+x+y=28 \\ -p-x-3y=-30 \\ ------------- \\ p-2y=-2 \end{gathered}$

Now, we can multiply the 3rd equation by (-1) and add it to the 2nd equation, we get:

$\begin{gathered} p+x+3y=30 \\ -3p-x-2y=-44 \\ -------------- \\ -2p+y=-14 \end{gathered}$

We have 2 new equations. We can multiply this last equation by 2 and add up these 2 new equations. We can solve for p:

$\begin{gathered} p-2y=-2 \\ 2\times(-2p+y=-14) \\ ------------- \\ p-2y=-2 \\ -4p+2y=-28 \\ ------------- \\ -3p=-30 \\ p=\frac{-30}{-3} \\ p=10 \end{gathered}$

We can take the last equation, put the value of p and find the value of y:

$\begin{gathered} p-2y=-2 \\ p=10, \\ 10-2y=-2 \\ 10+2=2y \\ 12=2y \\ y=\frac{12}{2} \\ y=6 \end{gathered}$

We take first equation (totally first one) and put in the values of p and y and solve for x:

$\begin{gathered} 2p+x+y=28 \\ 2(10)+x+6=28 \\ 20+x+6=28 \\ 26+x=28 \\ x=28-26 \\ x=2 \end{gathered}$

So,

x = 2

y = 6

p = 10

We can say:

one small pizza costs $10

one liter of pop costs $2

one salad costs $6