In basketball, there are 3-point shots, 2-point shots, and 1-point free throws. Josh scored 39 points in a game. He made the same number of 2-point shots as 3-point and free throws combined. He scored one more point on 3-pointers than he did on free throws. How many of each did he make?
Let
x1=number of 1-point shots
x2=number of 2-point shots
x3=number of 3point shots
"Josh scored 39 points in a game" =>
3x3+2x2+x1=39.........(1)
"the same number of 2-point shots as [3-point and free throws combined]"
x2=x3+x1 .............(2)
"scored one more point on 3-pointers than he did on free throws" =>
3x3-x1=1 .............(3)
Solve the system of equations:
Substitute (2) in (1)
3x3+2(x3+x1)+x1=39, simplify
5x3+3x1=39............(1a)
3 times (3):
9x3-3x1 =3 ............(3a)
(3a)+(1a)
14x3=42 => x3=3
Substitute x3 in (3)
3(3)-x1=1 => x1=8
substitute x3 and x1 in (1)
3(3)+2(x2)+8 = 39 =>
x2=(39-9-8)/2=11
So (x1,x2,x3)=(8,11,3)
Check answer by substituting values of x1,x2 and x3 into original equations.
2t + 3r + f = 39
t = r + f
3r - 1 = f
subs ... t = r + 3r - 1 = 4r - 1
subs ... 8r - 2 + 3r + 3r - 1 = 39
find r, then substitute back
Ah, I see we have a little basketball puzzle to solve here. Alright, let's put on our thinking hats, or should I say, our thinking headbands.
Let's assume Josh made x number of 3-point shots. That means he scored 3x points from those shots.
Now, since he scored one more point on 3-pointers than he did on free throws, we can say that he made (x - 1) free throws, which would give him (x - 1) points.
Moving on to 2-point shots, we're given that he made the same number of 2-point shots as the combined total of 3-pointers and free throws. So, that means he made a total of 2(x + (x - 1)) = 4x - 2 points.
Now, we can add up all the points: 3x + (x - 1) + (4x - 2) = 39.
Combining like terms, we have: 8x - 3 = 39.
Solving for x, we find that x = 6.
So, Josh made 6 3-point shots, (6 - 1) = 5 free throws, and 2(6 + (6 - 1)) = 20 2-point shots.
That's one sharpshooting clown, huh?
Let's solve this problem step-by-step:
Step 1: Let's assume Josh made x 3-point shots, y 2-point shots, and z 1-point free throws.
Step 2: According to the problem, Josh scored 39 points in total. We can write the equation as:
3x + 2y + z = 39
Step 3: The problem states that Josh made the same number of 2-point shots as the combined number of 3-point and 1-point free throws. We can write this as an equation:
y = x + z
Step 4: The problem also states that Josh scored one more point on 3-pointers than he did on free throws. We can write this equation as:
3x = z + 1
Step 5: Now we have a system of equations. We can solve this system using substitution or elimination method. Let's use substitution method:
From equation 3, we have z = 3x - 1.
Substituting this value of z in equation 2, we get:
y = x + (3x - 1)
y = 4x - 1
Substituting the value of z and y in equation 1, we get:
3x + 2(4x - 1) + (3x - 1) = 39
3x + 8x - 2 + 3x - 1 = 39
14x - 3 = 39
14x = 42
x = 3
Step 6: Now we can substitute the value of x back into equation 3 to find z:
z = 3(3) - 1
z = 8
Step 7: Finally, we can substitute the values of x and z into equation 2 to find y:
y = 4(3) - 1
y = 11
Step 8: So, Josh made 3 3-point shots, 11 2-point shots, and 8 1-point free throws.
Therefore, Josh made 3 3-point shots, 11 2-point shots, and 8 1-point free throws in the game.
To determine the number of each type of shot Josh made, let's use a system of equations:
Let's assume:
x = the number of 3-point shots made
y = the number of 2-point shots made
z = the number of 1-point free throws made
We're given the following information:
1. Josh scored 39 points:
3x + 2y + z = 39
2. He made the same number of 2-point shots as 3-point and free throws combined:
y = x + z
3. He scored one more point on 3-pointers than free throws:
x = z + 1
Now, we can solve these equations simultaneously to find the values of x, y, and z.
First, let's rearrange equation 2 to:
z = y - x
Substituting this value for z in equations 1 and 3, we get:
3x + 2y + (y - x) = 39 -> 4x + 3y = 39 -> equation 4
x = (y - x) + 1 -> 2x = y + 1 -> equation 5
Now we have a system with equations 4 and 5.
Let's solve equation 5 for x:
2x = y + 1 -> x = (y + 1)/2
Substitute this value for x in equation 4:
4x + 3y = 39 -> 4((y + 1)/2) + 3y = 39 -> 2(y + 1) + 3y = 39
2y + 2 + 3y = 39 -> 5y = 37 -> y ≈ 7.4
Since y represents the number of 2-point shots made, it must be a whole number. So, let's try y = 7.
Substituting y = 7 into equation 5:
x = (7 + 1)/2 -> x = 4
Substituting y = 7 and x = 4 into equation 3:
x = z + 1 -> 4 = z + 1 -> z = 3
Therefore, Josh made 4 three-pointers, 7 two-pointers, and 3 free throws to score 39 points in the game.