Question

a school is organizing a cookout where hotdogs will be served. the hotdogs come in small packs and large packs. each small pack has 6 hotdogs and each large pack has 12 hotdogs. the school bought twice as many small packs as large packs, which altogether had 96 hotdogs. graphically solve a system of equations in order to determine the number of small packs purchased, x, and the number of large packs purchased, y. click twice to plot each line. click a line to delete it.

Explanation:

Step1: Define variables and equations

Let \( x \) be the number of small packs and \( y \) be the number of large packs. We know two things:

1. The number of small packs is twice the number of large packs, so \( x = 2y \).
2. Each small pack has 6 hotdogs and each large pack has 12 hotdogs, and the total number of hotdogs is 96. So the equation for the total hotdogs is \( 6x + 12y = 96 \).

Step2: Substitute \( x = 2y \) into the second equation

Substitute \( x \) in \( 6x + 12y = 96 \) with \( 2y \):
\[
6(2y) + 12y = 96
\]
Simplify the left - hand side:
\[
12y+12y = 96
\]
\[
24y = 96
\]

Step3: Solve for \( y \)

Divide both sides of the equation \( 24y = 96 \) by 24:
\[
y=\frac{96}{24}=4
\]

Step4: Solve for \( x \)

Since \( x = 2y \) and \( y = 4 \), then \( x=2\times4 = 8 \).

To graphically solve the system:

• For the equation \( x = 2y \) (or \( y=\frac{1}{2}x \)):
• When \( x = 0 \), \( y = 0 \); when \( x = 4 \), \( y = 2 \); when \( x = 8 \), \( y = 4 \). Plot these points \((0,0)\), \((4,2)\), \((8,4)\) and draw the line.
• For the equation \( 6x + 12y=96 \), we can simplify it to \( x + 2y=16 \) (divide both sides by 6) or \( y=-\frac{1}{2}x + 8 \).
• When \( x = 0 \), \( y = 8 \); when \( x = 8 \), \( y = 4 \); when \( x = 16 \), \( y = 0 \). Plot these points \((0,8)\), \((8,4)\), \((16,0)\) and draw the line.

The intersection point of the two lines \( y=\frac{1}{2}x \) and \( y =-\frac{1}{2}x + 8 \) is \((8,4)\), which means \( x = 8 \) and \( y = 4 \).

Answer:

The number of small packs \( x = 8 \) and the number of large packs \( y = 4 \). So the solution to the system of equations (graphically, the intersection point) is \((8,4)\), where \( x = 8 \) (small packs) and \( y = 4 \) (large packs).