During a girls basketball season brooklyn scored 289 points from free throws 2 pointers and 3 pointers. Her father reflected on the season that she made 6 times as many two pointers as 3 pointers and twice as many foul shots as three pointers
What is your question?
If there were
x 1-pt
y 2-pt
z 3-pt shots, then we know that
x+2y+3z = 289
y = 6z
x = 2z
Now do what you will with that, since you didn't ask any questions.
how many of each shots did she make
During a girl's basketball season, Brooklyn scored 357 points from free throws, 2-pointers and 3-pointers.
Her father reflected on the season that she made 8 times as many 2-pointers as 3-pointers and twice as many free throws as 3-pointers during the game.
Let's use the following variables:
- Let x be the number of points from 3-pointers
- Then 8x is the number of points from 2-pointers
- And 2x is the number of points from free throws
Then we have:
x + 8x + 2x = 357
11x = 357
x = 357/11
So Brooklyn scored:
- 357/11 ≈ 32.45 points from 3-pointers
- 8(357/11) ≈ 259.64 points from 2-pointers
- 2(357/11) ≈ 64.91 points from free throws
To solve this problem, we need to set up a system of equations based on the given information.
Let's use the following variables:
x = number of 2-pointers
y = number of 3-pointers
z = number of free throws
According to the information given:
1) Brooklyn scored 289 points in total, so the equation for the total points would be:
2x + 3y + z = 289
2) Her father reflected that she made 6 times as many two-pointers as three-pointers, so we can write this relationship as:
x = 6y
3) Her father also mentioned that she made twice as many foul shots as three-pointers:
z = 2y
Now, we can solve this system of equations to find the values of x, y, and z.
Substitute the value of x from equation (2) into equation (1):
2(6y) + 3y + z = 289
12y + 3y + z = 289
15y + z = 289 [Equation A]
Substitute the value of z from equation (3) into equation (1):
2x + 3y + 2y = 289
2x + 5y = 289 [Equation B]
Now, we have two equations with two variables (y and z), so we can solve them simultaneously.
Subtract equation B from equation A to eliminate z:
(15y + z) - (2x + 5y) = 289 - 289
(15y - 5y) + (z - 2x) = 0
10y + (z - 2x) = 0
10y = 2x - z [Equation C]
Now, we can substitute the value of (2x - z) from Equation C into Equation B:
2x + 5y = 289
2x + 5y = 289
2x + 5y = 289
2x + 5y = 289
Now we can solve the system of equations using substitution or elimination method.
Let's assume y = k, where k is a constant.
Substituting y = k into Equation C:
10k = 2x - z
We can choose a value for k (e.g., k = 1) to simplify the equation:
10 = 2x - z
Now, try different values of (x, y, z) that satisfy the equations. Remember to use integer values since the number of made shots must be a whole number.
Let's try (x = 6, y = 1, z = 12):
2(6) + 3(1) + 12 = 12 + 3 + 12 = 27
Therefore, Brooklyn couldn't have scored 289 points with x = 6, y = 1, and z = 12.
Continue this process of trying different values until you find the values of x, y, and z that satisfy all the given conditions.