Q. 80 - Precalculus Enhanced with Graphing Utilities

Question

Use the information given in Problem 79. Suppose that a third group purchased 3 deluxe hamburgers, 2 large fries, and 4 large colas for $10.95. Now is there sufficient information to determine the price of each

food item? If so, determine each price.

Step-by-Step Solution

Answer

The prices of deluxe hamburgers, large fries and colas are 1.95, 0.95, 0.8 respectively.

Step 1: Given information

a third group purchased 3 deluxe hamburgers, 2 large fries, and 4 large colas for $10.95.  

Step 2: Find the equation

Let x represent the hamburgers

Let y represent the fries

let z represent the colas

For the first group we can write

$8x+6y+6z=26.10 \left(1\right)$

For the second group we can write

$10x+6y+8z=31.60 \left(2\right)$

For the third group we write

$3x+2y+4z=10.95$

Step 3: Solve the equations

Multiply equation 1 by -1 and add it to equation 2 we get

$-8x-6y-6z=-26.1+10x+6y+8z=31.62x+2z=5.5$

We get,$2x+2z=5.5$     

 $x+z=2.75 \left(4\right)$

Now,

Multiply equation 3 by -3 and add it to equation 1 we get

$8x+6y+6z=26.1-9x-6y-12z=-32.85-x-6z=-6.75$

We get, $-x-6z=-6.75$                (5)

Step 4: Solve the equation in two variables

We get,

$x+z=2.75-x-6z=-6.75-5z=-4$

Hence $z=0.8$

Substitute this in equation 4

$$\begin{gathered} x+z=2.75 \\ x+0.8=2.75 \\ x=1.95 \end{gathered}$$

Now find the value of y

$$\begin{gathered} 8x+6y+6z=26.1 \\ 8(1.95)+6y+6(0.8)=26.1 \\ 6y=26.1-15.6-4.8 \\ 6y=5.7 \\ y=0.95 \end{gathered}$$

Step 5: Conclusion

The prices of deluxe hamburgers, large fries and colas are 1.95, 0.95, 0.8 respectively.