Question

A

10. In one basketball game, a player scored
31 points with a combination of two-point
baskets, three-point baskets, and one-
point free throws. She made 5 more two-
point baskets than one-point free throws,
and 3 times as many two-point baskets
as three-point baskets. How many three-
point baskets, two-point baskets, and
free throws did the player make?

Answer 1

ok

Explanation:

Answer 2

• 1-point: 4
• 2-point: 9
• 3-point: 3

Explanation:

Let x, y, z represent the numbers of 1-, 2-, and 3-point baskets, respectively. The given relations are ...

x +2y +3z = 31 . . . . . . total points scored

y -x = 5 . . . . . . . . . . . 5 more 2-point baskets than free throws

y = 3z . . . . . . . . . . . . 3 times as many 2-point baskets as 3-point baskets

Using the second equation to write an expression for x, we have ...

x = y -5

Substituting this and the third equation into the first, we get ...

(y -5) +2y +(y) = 31

4y = 36 . . . . . . . . . . . add 5, collect terms

y = 9

x = 9 -5 = 4

z = y/3 = 9/3 = 3

The player made 4 free throws, 9 2-point baskets, and 3 3-point baskets.

_____

Alternate solution

The equations can be written in augmented matrix form as ...

`\left[\begin{array}ccc1&2&3&31\\-1&1&0&5\\0&1&-3&0\end{array}\right]`

Row-reducing this matrix using any of a number of calculators gives (x, y, z) = (4, 9, 3), as above.