Q30. In the 2000-2001 season, Minneso... [FREE SOLUTION]

Chapter 3: Q30. (page 143)

In the 2000-2001 season, Minnesota鈥檚 Katie Smith was ranked first in the WNBA for total points and three-point field goals made. She scored 646 points making 355 shots, including 3-point field goals, 2-point field goals, and 1-point free throws. She made 27 more 2-point field goals than 3- point field goals.
Find the number of each type of goal she made.

Short Answer

Expert verified

She made 88 three-point field goals, 115 two-point field goals, and 152 one point free throws.

Step by step solution

– First construct a system of equations corresponding to the given situation.

Let $x$ be the number of 1-point free throws, $y$ be the number of 2-point field goals, and $z$ be the number of 3-point field goals.

In the figure, it is given that the total 355 shots are made.

So, $x + y + z = 355$ .

In the figure, it is given that the total scored points is 646.

So, $x + 2 y + 3 z = 646$

In the figure, it is given that the 27 more 2-point field goals than 3-point field goals.

This can be mathematically written as $y = z + 27 鈬� y - z = 27$ .

So, the system of equations will be:

$\begin{array}{c} x + y + z = 355 \\ x + 2 y + 3 z = 646 \\ y 鈭� z = 27 \end{array}$

– Use the elimination method to get the system of equations in two variables.

Add $x + y + z = 355$ to the equation $y - z = 27$ .

$\begin{array}{l} \underline{\begin{array}{l} x + 鈥夆赌夆赌� y + z = 355 \\ 鈥夆赌夆赌夆€夆赌夆赌夆€� y 鈭� z = 鈥夆赌夆赌� 27 \end{array}} \\ x + 2 y + 0 = 382 \end{array}$

So, the resultant equation is $x + 2 y = 382$

Multiply the $e q u a t i o n y - z = 27$ by 3 and add the new resultant equation to the equation $x + 2 y + 3 z = 646$

$\begin{array}{l} \underline{\begin{array}{l} 鈥夆赌夆赌夆€夆赌夆赌夆€� y 鈭� 鈥夆赌� z = 鈥夆赌夆赌� 27 \\ x + 2 y + 3 z = 646 \end{array}} 鈥夆赌� \begin{matrix} multiplyby3 \end{matrix} 鈥夆赌夆赌� \underline{\begin{array}{l} 鈥夆赌夆赌夆€夆赌� 3 y 鈭� 3 z = 鈥夆赌夆赌� 81 \\ x + 2 y + 3 z = 646 \end{array}} \\ 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€� x + 5 y + 鈥夆赌夆€� 0 = 727 \end{array}$

.So, the resultant equation is

x + 5 y = 727

.
.

– Solve the system of equations in two variables

Subtract $x + 2 y = 382$ from $x + 5 y = 727$ as follows:

$\begin{array}{l} \underline{\begin{array}{l} 鈥夆赌夆赌夆€� x + 5 y = 727 \\ (鈭�) x + 鈥� 2 y = 382 \end{array}} \\ 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌� 3 y = 345 \end{array}$

Solve $3 y = 345$ for $y$ :

$\begin{matrix} 3 y = 345 \\ \frac{3 y}{3} = \frac{345}{3} 鈥夆赌夆赌夆€夆赌塪ividebothsidesby 3 \\ y = 115 \end{matrix}$

– Find the values of x and z.

Substitute $y = 115 i n x + 2 y = 382$ and find the value of $x$

$\begin{matrix} x + 2 y = 382 \\ x + 2 (115) = 382 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌塖ubstitute 115 for y \\ x + 230 = 382 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌塖implify \\ x = 152 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€塖ubtract 230 frombothsides \end{matrix}$

Substitute $y = 115 i n y - z = 27$ and find the value of $z$

$\begin{matrix} y 鈭� z = 27 \\ 115 鈭� z = 27 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌塖ubstitute 115 for y \\ 鈭� z = 鈭� 88 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌塖ubtract 115 frombothsides \\ z = 88 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌塂ividebothsidesby-1 \end{matrix}$

.Hence, the solution of the given system of equations is

(x, y, z) = (152, 115, 88)

.So, she made 88 three-point field goals, 115 two-point field goals, and 152 one point free throws.

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Short Answer

Step by step solution

– First construct a system of equations corresponding to the given situation.

– Use the elimination method to get the system of equations in two variables.

– Solve the system of equations in two variables

– Find the values of x and z.

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