Answer to Question #189295 in Algebra for Micah

Question

At the arcade, Sammy won two blue tickets, 1 yellow ticket and 3 red tickets for a total of 1,500 points. Jamal won 1 blue, 2 yellow and 2 red for 1225 points. Yvonne won 2 blue, 3 yellow and 1 red for 1200 points. How much is each ticket worth?

Answer 1

Let x, y, z be the points for blue, yellow, and red tickets respectively.

For Sammy,

2x + y + 3z = 1500 => (1)

For Jamal,

x + 2y + 2z = 1225 => (2)

For Yvonne,

2x + 3y + z = 1200 => (3)

Looking at equation (1),

2x + y + 3z = 1500

i.e., y = 1500 - 2x - 3z => (4)

Substituting this value of y in equations (2) & (3).

Equation (2) becomes,

x + 2(1500 - 2x - 3z) + 2z = 1225

i.e., x + 3000 - 4x - 6z + 2z = 1225

i.e., 3000 - 3x - 4z = 1225

i.e., 3x + 4z = 3000 - 1225

i.e., 3x + 4z = 1775 => (5)

Similarly, equation (3) becomes,

2x + 3(1500 - 2x - 3z) + z = 1200

i.e., 2x + 4500 - 6x - 9z + z = 1200

i.e., 4500 - 4x - 8z = 1200

i.e., 4x + 8z = 4500 - 1200

i.e., 4(x + 2z) = 3300

i.e., x + 2z = 3300/4 = 825

$\therefore$ x = 825 - 2z

Substituting the value x = 825 - 2z in equation (5),

3(825 - 2z) + 4z = 1775

2475 - 6z + 4z = 1775

2475 - 1775 - 2z = 0

2z = 2475 - 1775

2z = 700

$\therefore$ z = 350

$\therefore$ x = 825 - 2z = 825 - 700

i.e., x = 125

$\therefore$ y = 1500 - 2x - 3z

i.e., y = 1500 - 2(125) - 3(350)

i.e., y = 1500 - 250 - 1050

$\therefore$ y = 200

$\therefore$ The blue, yellow, and red tickets are worth 125, 200, and 350 points respectively.