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Question

 a basketball team recently scored a total of 78 points on a combination of a 2 point  field goals; 3  point field goals, and 1 point foul shots. Altogether, the team made 43 baskets and 15 more 2 -pointers than foul shouts. How many shots of each kind were made?

 

1 point =

2- point=

3- point=

please help provide answers Cleary!!!

 

 

Answer 1

Let x = the number of 1 pt shots made

Then x + 15 = the number of 2 pt shots made

Since there were a total of 43 baskets made, then the # of 3 pt shots is equal to:

 

total number of shots made minus the sum of 1 and 2 pt shots, which algebraically speaking is

 

43 - (x + (x + 15)) = 43 - (2x + 15) = 28 - 2x = # of 3 pt shots made

 

Now, we can set up the following equations:

 

x(1) + (x + 15)(2) + (28 - 2x)(3) = 78 (the total number of points)

 

x + 2x + 30 + 84 - 6x = 78

 

3x + 114 - 6x = 78

 

114 - 3x = 78

 

-3x = -36

 

x = 12

 

So, there were 12 one point shots made

 

(x + 15) = 12 + 15 = 27 two point shots made

 

and 

 

28 - 2x = 28 - 2(12) = 28 - 24 = 4 three pt shots made