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?

Find the number of three-point baskets, two-point baskets, and free throws did the player make:

Given that 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.

Assuming that the player scored $\text{'}\mathrm{x}\text{'}$ number of two point baskets ,$\text{'}\mathrm{y}\text{'}$ number of three point baskets and $\text{'}\mathrm{z}\text{'}$ denotes the free throws.

Player scored $31$ points with a combination of two-point baskets, three-point baskets, and one free throw:

$\Rightarrow 2\mathrm{x}+3\mathrm{y}+\mathrm{z}=31...\left(1\right)$

She scored $5$ more two-point baskets than one-point free throws:

$\begin{array}{rcl}& \Rightarrow & \mathrm{x}=\mathrm{z}+5\\ & \Rightarrow & \mathrm{z}=\mathrm{x}-5...\left(2\right)\end{array}$

She scored $3$ times as many two-point baskets as three-point baskets:

$\Rightarrow \mathrm{x}=3\mathrm{y}...\left(3\right)$

Step-1: Find the value of $\mathrm{y}$:

Substitute $\mathrm{z}=\mathrm{x}-5$ in equation (1):

$\begin{array}{rcl}& \Rightarrow & 2\mathrm{x}+3\mathrm{y}+\mathrm{x}-5=31\\ & \Rightarrow & 3\mathrm{x}+3\mathrm{y}=31+5\\ & \Rightarrow & 3\left(\mathrm{x}+\mathrm{y}\right)=36\\ & \Rightarrow & \left(\mathrm{x}+\mathrm{y}\right)=\frac{36}{3}\\ & \Rightarrow & \mathrm{x}+\mathrm{y}=12...\left(4\right)\end{array}$

Now, substitute $\mathrm{x}=3\mathrm{y}$ in equation $\left(4\right)$:

$\begin{array}{rcl}& \Rightarrow & 3\mathrm{y}+\mathrm{y}=12\\ 4\mathrm{y}& =& 12\\ \mathrm{y}& =& 3\end{array}$

So, the number of three-point baskets are $3$.

Step-2: Find the value of $\mathrm{x}$.

Substitute $\mathrm{y}=3$ in equation $\left(4\right)$:

$\begin{array}{rcl}& \Rightarrow & \mathrm{x}+3=12\\ & \Rightarrow & \mathrm{x}=12-3\\ & \Rightarrow & \mathrm{x}=9\end{array}$

So, the number of two-point baskets are $9$.

Step-3: Find the value of $\mathrm{z}$.

Substitute $\mathrm{x}=9$ in equation $\mathrm{z}=\mathrm{x}-5$:

$\begin{array}{rcl}& \Rightarrow & \mathrm{z}=9-5\\ & \Rightarrow & \mathrm{z}=4\end{array}$

So, the number of free throws are $4$.

Hence, the number of three-point baskets, two-point baskets, and free throws the player made is $\mathbf{3}\mathbf{,}\mathbf{9}\mathbf{,}\mathbf{4}$ respectively.