In a competition 3 points are awarded for every game won and 2 points are deducted for every game lost.A pupil attempted 5 games and got 5 points.How many games did he lose?
x = number of game won
y = number of game lost
A pupil attempted 5 games means:
x + y = 5
He got x ∙ 3 points for the games he won.
For lost games he lost y ∙ 2 points.
He got 5 points means:
x ∙ 3 - y ∙ 2 = 5
3 x - 2 y = 5
Now you must solve system of two equations;
x + y = 5
3 x - 2 y = 5
The solution is:
x = 3 , y = 2
He won 3 games
He lose 2 games
Check result.
x + y = 3 + 2 = 5
He attempted 5 games.
He got for winning the games x ∙ 3 = 3 ∙ 3 = 9 points
He lost for lost games 2 ∙ 2 points 2 = 4 points
9 points - 4 points = 5 points
3w-2(5-w)=5
3w-10+2w=5
5w_10=5
5w=15
W=3
Games won =3
Games lost=2
Correct! The pupil won 3 games and lost 2 games.
To solve this problem, we need to set up an equation. Let's assume that the number of games won is represented by 'x' and the number of games lost is represented by 'y'.
According to the given information, 3 points are awarded for every game won, and 2 points are deducted for every game lost. So, the total points earned can be calculated as:
3x - 2y = 5 ...(Equation 1)
The other piece of information given is that the pupil attempted a total of 5 games, which means the sum of games won and games lost should be equal to 5:
x + y = 5 ...(Equation 2)
We now have a system of two linear equations with two variables. We can solve this system to find the values of 'x' and 'y', representing the number of games won and lost, respectively.
One way to solve this system is by substitution. We can rearrange Equation 2 to express 'x' in terms of 'y':
x = 5 - y
Now we substitute this expression for 'x' in Equation 1:
3(5 - y) - 2y = 5
Distribute the 3:
15 - 3y - 2y = 5
Combine like terms:
15 - 5y = 5
Rearrange the equation:
-5y = 5 - 15
-5y = -10
Divide by -5:
y = -10 / -5
y = 2
So, the pupil lost 2 games.