SOLUTION: I need help with this word problem during the 2003 NBA playoffs, the top-scoring player was Kobe Bryant of the Los Angeles Lakers. Bryant scored a totals of 385 points during

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Bryant scored a totals of 385 points during the playoffs. The number of free throws (each worth one point) he made was 11 more than three times the number of three-point field goals he made. Bryant also made 26 fewer free throws than two-point field goals. How many free throws, two point field goals. and three point field goals did Kobe Bryant make during the 2003 playoffs?
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EQUATIONS:
f = 3r + 11
f = t - 26
f + 2t + 3r=385
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Rearrange:
f + 2t + 3r = 385
f + 0t - 3r = 11
f - t + 0r = -26
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Solve the system of equations by any method to get:
# of free throws = 86
# of two-pointers = 112
# of three-pointers = 25
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Cheers,
Stan H.

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Let = the number of free throws he made
Let = the number of three-point shots he made
Let = the number of two-point shots he made
The problem says
(1)
and
(2)
(3)
----------------
(2)
Solve for


----------------
(3)
Solve for

----------------
Now plug the expressions for and
back into (1)
(1)





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And since,
(2)



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Also,
(3)


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Kobe made freethrows, 3-pointers,
and 2-point shots
Check the answer:
(1)
(1)


OK