System of Equations and Matrices

[TiaharaJBennett]

Hello!

Okay, so I need some help. I have these word problems I have to do for homework, like this:

The Arcadium ardcade in Lynchburg, Tennessee uses 3 different colored tokens for their game machines. For 20$ you can purchase any of the following mixtures of tokens: 14 gold, 20 silver, and 24 bronze; OR, 20 gold, 15 silver, and 19 bronze; OR, 30 gold, 5 silver, and 13 bronze.

These problems tell me to write a system of equations (and I did!), which is:

14x+20y+24z=20
20x+15y+19z=20
30x+5y+13z+20

x represents the value of gold
y represents the value of silver
z represents the value of bronze

I then had to represent the system as a matrix, which is:

14 20 24 x 20
20 15 19 y 20
30 5 13 z 20

All I want to know is how do I figure out the monetary value of each token? Could someone explain to me how I do that?

Thanks.

 

[Deleted member 4993]

Hello!

Okay, so I need some help. I have these word problems I have to do for homework, like this:

The Arcadium ardcade in Lynchburg, Tennessee uses 3 different colored tokens for their game machines. For 20$ you can purchase any of the following mixtures of tokens: 14 gold, 20 silver, and 24 bronze; OR, 20 gold, 15 silver, and 19 bronze; OR, 30 gold, 5 silver, and 13 bronze.

These problems tell me to write a system of equations (and I did!), which is:

14x+20y+24z=20
20x+15y+19z=20
30x+5y+13z+20

x represents the value of gold
y represents the value of silver
z represents the value of bronze

I then had to represent the system as a matrix, which is:

14 20 24 x 20
20 15 19 y 20
30 5 13 z 20

All I want to know is how do I figure out the monetary value of each token? Could someone explain to me how I do that?

Thanks.

For a quick review - please go to:

http://www.purplemath.com/modules/systlin6.htm

 

[HallsofIvy]

Why did you write it as a matrix? That is a perfectly good method but the fact that you mention "matrices" makes me think you must know something about them!

You have \(\displaystyle \begin{bmatrix}14 & 20 & 24 \\ 20 & 15 & 19 \\ 30 & 5 & 13\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}20 \\ 20 \\ 20\end{bmatrix}\)
Writing it like that, the obvious thing to do is to find the inverse matrix of the coefficient matrix, then multiply both sides by that. That is, you solve Ax= b by multiplying both sides by \(\displaystyle A^{-1}\): \(\displaystyle A^{-1}Ax= x= A^{-1}x\),

Another way to solve a systme of equations like that is to write the "augmented matrix":
\(\displaystyle \begin{bmatrix}14 & 20 & 24 & 20\\ 20 & 15 & 19 & 20 \\ 30 & 5 & 13 & 20 \end{bmatrix}\)
and "row reduce" so that the first three columns are \(\displaystyle \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\) and the last column will give x, y, and z.

But I will admit that I, personally would not use "matrices" at all. From the equation \(\displaystyle 14x+ 20y+ 24z= 20\) we can divide by 2 to get \(\displaystyle 7x+ 10y+ 12z= 10\). The second equation is \(\displaystyle 20x+ 15y+ 19z= 20\). If we multiply that equation by 2, the first equation by 3, and subtract the second from the first, we get \(\displaystyle (40x+ 30y+38z)- (21x+ 30y+ 36z)= 60-20\) or \(\displaystyle 19x+ 2z= 40\), eliminating y. The subtract \(\displaystyle 7x+ 10y+ 12x= 10\) from, say, twice the third equation, \(\displaystyle 60x+ 10y+ 26z= 40\), to get \(\displaystyle 53x+ 14z= 30\). Now we have two equations in two unknowns. Manipulate those equations to eliminate one of those.