A basketball team recently scored a total of 92 points on a combination of 2-point field goals, 3-point field goals, and 1-point foul shots. Altogether, the team made 51 baskets and 22 more 2-pointers than foul shots. How many shots of each kind were made?
If there are x 1-pts, y 2-pts and z 3-pts, then we have
x+2y+3z = 92
x+y+z = 51
y = x+22
Now just grind it out.
Freud
Let's solve this step by step.
Let's assume the number of 1-point foul shots made is x.
According to the given information, the number of 2-point field goals made would be x + 22 (since there are 22 more 2-pointers than foul shots).
Also, the number of 3-point field goals made would be 51 - (x + (x + 22)) = 29 - 2x.
Next, we can calculate the total points scored:
1-point foul shots: x points
2-point field goals: (x + 22) * 2 points
3-point field goals: (29 - 2x) * 3 points
According to the problem, the total points scored is 92, so we can set up the equation:
x + 2(x + 22) + 3(29 - 2x) = 92
Now, let's solve the equation step by step:
x + 2x + 44 + 87 - 6x = 92
-3x + 131 = 92
-3x = -39
x = 13
Therefore, the number of 1-point foul shots made is 13.
The number of 2-point field goals made would be x + 22 = 13 + 22 = 35.
The number of 3-point field goals made would be 29 - 2x = 29 - 2(13) = 3.
So, the team made 13 1-point foul shots, 35 2-point field goals, and 3 3-point field goals.
To solve this problem, let's define some variables:
Let's say the number of 2-point field goals made is x.
Let's say the number of 3-point field goals made is y.
Let's say the number of 1-point foul shots made is z.
Based on the information given, we can form equations:
1. The total number of baskets made: x + y + z = 51
2. The total number of points scored: 2x + 3y + z = 92
3. The number of 2-pointers is 22 more than the number of foul shots: x = z + 22
Now we have a system of three equations that we can solve to find the values of x, y, and z.
Let's start solving this system of equations:
From equation 3, we can substitute x in terms of z in the other equations:
1. z + 22 + y + z = 51 -> 2z + y = 29
2. 2(z + 22) + 3y + z = 92 -> 3z + 3y = 48 -> z + y = 16
Now we have a system of two equations with two variables. We can solve this system using substitution or elimination.
Let's use substitution:
From equation 1, we can express y in terms of z:
y = 16 - z
Now substitute this into equation 2:
z + (16 - z) = 29
2z + 16 = 29
2z = 13
z = 6.5
Now substitute the value of z back into one of the other equations to find the values of x and y.
From equation 3:
x = z + 22
x = 6.5 + 22
x = 28.5
From equation 1:
x + y + z = 51
28.5 + y + 6.5 = 51
y + 35 = 51
y = 16
So the number of 2-point field goals made is 28.5, the number of 3-point field goals made is 16, and the number of 1-point foul shots made is 6.5.
Since we can't have half shots, we can round down the 6.5 to 6.
Therefore, the basketball team made 28 two-point field goals, 16 three-point field goals, and 6 one-point foul shots.