Problema Solution
You and your best friend are playing Call of Duty. He has two more kills than you in the first game. In the second game you have 10 less than twice the kills you had the first game, and your friend has fifteen less than three times the amount he had his first game. In the second game played you have 16 less than him in kills. How many kills did you both have in the first games
Answer provided by our tutors
First game:
x = the number of kills you have in the first game
y = the number of kills your friend has in the first game
He has two more kills than you in the first game:
y = x + 2
Second game:
a = the number of kills you have in the second game
b = the number of kills your friend has in the second game
In the second game you have 10 less than twice the kills you had the first game:
a = 2x - 10
and your friend has fifteen less than three times the amount he had his first game:
b = 3y - 15
In the second game played you have 16 less than him in kills:
a = b - 16
plug a = 2x - 10 and b = 3y - 15 into a = b - 16:
2x - 10 = 3y - 15 - 16
2x = 3y - 21
by solving the system of equations:
y = x + 2
2x = 3y - 21
we find:
x = 15 kills in the first game
y = 17 kills in the first game
click here to see the step by step solution of the system of equations:
In the First game you had 15 kills and your friend had 17 kills.