NBA Records. The three highest one-game point totals by one...

NBA Records. The three highest one-game point totals by one player in a National Basketball Association game are shown below. Solve the following system to find $x, y,$ and $z$ $$ \left\{\begin{array}{l} x+y+z=259 \\ x-y=19 \\ x-z=22 \end{array}\right. $$ $$ \begin{array}{|c|l|c|} \hline \text { Pts } & \text { Player, team } & \text { Date } \\ \hline x & \text { Wilt Chamberlain, Philadelphia } & 3 / 2 / 1962 \\ y & \text { Kobe Bryant, Los Angeles } & 1 / 22 / 2006 \\ z & \text { Wilt Chamberlain, Philadelphia } & 12 / 8 / 1961 \\ \hline \end{array} $$

NBA Records. The three highest one-game point totals by one player in a National Basketball Association game are shown below. Solve the following system to find $x, y,$ and $z$
$$
\left\{\begin{array}{l}
x+y+z=259 \\
x-y=19 \\
x-z=22
\end{array}\right.
$$
$$
\begin{array}{|c|l|c|}
\hline \text { Pts } & \text { Player, team } & \text { Date } \\
\hline x & \text { Wilt Chamberlain, Philadelphia } & 3 / 2 / 1962 \\
y & \text { Kobe Bryant, Los Angeles } & 1 / 22 / 2006 \\
z & \text { Wilt Chamberlain, Philadelphia } & 12 / 8 / 1961 \\
\hline
\end{array}
$$

Key Concepts

Elimination Method

The elimination method involves combining equations in a way that removes one or more variables. By adding or subtracting equations, one can cancel out a variable, thereby reducing the system to fewer equations and making it easier to solve for the remaining variables.

Systems of Linear Equations

Systems of linear equations involve finding the set of variable values that satisfy all given linear equations simultaneously. In the context of the problem, understanding how multiple equations relate to the same unknowns is crucial for determining a unique solution that works for all equations.

Substitution Method

The substitution method is an algebraic technique used to solve systems of equations by isolating one variable in one equation and substituting its expression into the other equations. This method simplifies the system, reducing the number of variables and enabling step-by-step solving.

Transcript

00:01 Okay, so this is all about national basketball association.

00:06 And each of these variables, x, y, and z, represents a player who scored the most game points in one game.

00:20 We had x represents, we had wilton chamberlain in 1962, scoring quite high.

00:31 Y, being kobe bryant, 2006, and z, again, being wilts chamberlain, but in 1962.

00:46 So even though x and z are the same person, we have to consider them two different variables because he scored, these are the top scoring individuals who scored the highest, points in one game and so wilts chamberman chamberlain did it twice right two different years which wow what an amazing accomplishment i wish i had that in my lifetime so we are going to look at this and see we're going to solve for the number of points that they scored in that one game so notice my equation two and my equation three only has two variables and they both have x.

01:36 So i'm going to use substitution in order to solve the system.

01:42 So i'm going to solve for y in this first equation, leaving x by itself.

01:50 But what i'm going to do, and i hope i don't confuse you, is i am going to move y to the right side and 19 back to the left side, just because it's a negative y, i want it to be a positive y.

02:07 So when i move y to the right side, i'm going to add y to both sides, giving me x equals 19 plus y, and then moving 19 back to the, or not back, but moving 19 to the left side, i'm going to subtract it, giving me x subtract 19 gives me positive y.

02:30 Now, i could have moved x over and then distributed, and then moved the negative 1 over.

02:36 This is how i chose to do it.

02:38 I'll do that for this third equation here.

02:41 Again, i'm going to solve it for z because i want those same variables.

02:47 Okay, so i'm going to, i'll do it the other way, just so you can see both ways.

02:53 This time i'm going to move x to the other side, giving me a negative z equals 22 subtract x.

03:02 This negative z is a negative 1 z.

03:05 Z.

03:05 Now i need to move that negative 1 to the other side by dividing, leaving me with a positive z, a negative 22, and a positive x.

03:20 So see, it's the same thing, just two different steps, but giving you the same result.

03:29 Now i'm going to go back up to equation 1 and plug in what i know is equal to y, which is x subtract 19.

03:39 And plug in what i know is equal to z which is negative 22 plus x now the that's a 9 um the y and the z both didn't have a coefficient in front of them so i have no distributing so i don't necessarily need those um parentheses so i'm just going to get rid of them and now i'm going to combine like terms so on my left side i have one x plus another one x plus another one x plus another one x giving me 3x total.

04:15 I have a negative 19 plus a negative 22, giving me a negative 41, and then i still have that equal to 259...

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