Answer:
The correct option is (c).
Explanation:
Let x and y denote the cost of game 1 and game 2 respectively.
ATQ,
4x+6y = 18 ....(1)
5x + 5y = 20 ....(2)
Multiply equation (1) by 5 and equation (2) by 4.
20x+30y = 90 ....(3)
20x + 20y = 80 ....(4)
Subtracting equation (3) and (4)
20x+30y - (20x + 20y) = 90-80
20x-20x+30y-20y = 10
10y = 10
y = 1
Put the value of y in equation (1).
4x+6(1) = 18
4x = 18-6
4x = 12
x = 3
Hence, the cost of game 1 is $3.00