Question

Jackson went into a movie theater and bought 8 bags of popcorn (x) and 2 candies (y), costing a total of $58. Cooper went into the same movie theater and bought 4 bags of popcorn (x) and 5 candies (y), costing a total of $49. Write a system of equations that could be used to determine the price of each bag of popcorn and the price of each candy. What is the price of 1 bag of popcorn and 1 candy?

Answer 1

Answer:

(a)

$8x + 2y = 58$

$4x + 5y = 49$

(b)

Bag of popcorn = $6

Candy = $5

Step-by-step explanation:

Given

$candy = y$

$popcorn = x$

Jackson: x = 8; y= 2; Total = 58

Cooper: x = 4; y= 5; Total = 49

Solving (a): System of equation

For Jackson:

$8x + 2y = 58$

This is done by multiplying each variable by the quantity.

Same will be applied to Cooper's purchase:

Cooper:

$4x + 5y = 49$

Solving (b): Price of 1 bag of popcorn and 1 candy

This implies that we solve for x and y

$8x + 2y = 58$ ----- (1)

$4x + 5y = 49$ -------- (2)

Multiply the second equation by 2

$2(4x + 5y = 49)$

$8x + 10y = 98$ -------- (3)

Subtract the (1) from (3)

$8x - 8x + 10y - 2y = 98 - 58$

$10y - 2y = 98 - 58$

$8y = 40$

Solve for y

$y = 40/8$

$y = 5$

Substitute 5 for y in (1)

$8x + 2y = 58$

$8x + 2 * 5 = 58$

$8x + 10 = 58$

$8x = 58 - 10$

$8x = 48$

Solve for x

$x = 48/8$

$x = 6$

This implies:

Bag of popcorn = $6

Candy = $5