Solve Word Problem w/ Matrices: Chapter I

[megacat8921]

I have a word problem that I am struggling with. I have been using matrices in this chapter, but I don't understand how it applies or where to start in order to solve this equation. Here is the word problem:

One hundred liters of a 50% solution is obtained by mixing a 60% solution with a 20% solution. How many liters of each solution must be used to obtain the desired mixture?

 

[MarkFL]

I would let $x$ represent the amount (in liters) of the 60% solution needed, and $y$ be the number of liters of the 20% solution required. Since the final desired outcome is 100 liters of solution, we know:

\(\displaystyle x+y=100\)

We also know that we will need in the final solution 50L of the active ingredient, $0.6x$ coming from the 60% solution and $0.2y$ coming from the 20% solution, then we also have:

\(\displaystyle 0.6x+0.2y=50\)

or:

\(\displaystyle 3x+y=250\)

So, we can set up our matrix equation as follows:

\(\displaystyle \left[\begin{array}{c}1 & 1 \\ 3 & 1 \end{array}\right]\left[\begin{array}{c}x \\ y \end{array}\right]=\left[\begin{array}{c}100 \\ 250 \end{array}\right]\)

Can you proceed?

 

[megacat8921]

I can proceed. But I don't understand how .6x and .2y became 3 and 1.

I would let $x$ represent the amount (in liters) of the 60% solution needed, and $y$ be the number of liters of the 20% solution required. Since the final desired outcome is 100 liters of solution, we know:

\(\displaystyle x+y=100\)

We also know that we will need in the final solution 50L of the active ingredient, $0.6x$ coming from the 60% solution and $0.2y$ coming from the 20% solution, then we also have:

\(\displaystyle 0.6x+0.2y=50\)

or:

\(\displaystyle 3x+y=250\)

So, we can set up our matrix equation as follows:

\(\displaystyle \left[\begin{array}{c}1 & 1 \\ 3 & 1 \end{array}\right]\left[\begin{array}{c}x \\ y \end{array}\right]=\left[\begin{array}{c}100 \\ 250 \end{array}\right]\)

Can you proceed?

 

[MarkFL]

I can proceed. But I don't understand how .6x and .2y became 3 and 1.

I multiplied the equation by 5 so that all coefficients are integers. :D