Asked by Nae
                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?
            
            
        Answers
                    Answered by
            Mathmate
            
    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 => <b>x3=3</b>
Substitute x3 in (3)
3(3)-x1=1 => <b>x1=8</b>
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.
    
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 => <b>x3=3</b>
Substitute x3 in (3)
3(3)-x1=1 => <b>x1=8</b>
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.
                    Answered by
            Scott
            
    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
    
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
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