Solve Word Problem w/ Matrices: Chapter I

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Discussion Overview

The discussion revolves around solving a word problem involving the mixing of two solutions to achieve a desired concentration using matrices. Participants explore how to set up the equations and matrix representation based on the problem's requirements.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents the word problem and seeks guidance on applying matrices to solve it.
  • Another participant proposes a system of equations based on the problem, defining variables for the amounts of each solution and establishing equations for total volume and active ingredient concentration.
  • A third participant expresses confusion regarding the transformation of the equation involving the active ingredient from $0.6x + 0.2y = 50$ to $3x + y = 250$.
  • A later reply clarifies that the transformation was achieved by multiplying the equation by 5 to eliminate decimals.

Areas of Agreement / Disagreement

Participants generally agree on the setup of the problem and the equations involved, but there is some confusion regarding the manipulation of the equations, particularly the conversion of coefficients.

Contextual Notes

There is an assumption that participants are familiar with matrix operations and the context of the problem, but the specific steps in manipulating the equations remain unclear to some.

Who May Find This Useful

Students or individuals interested in applying matrices to solve real-world problems, particularly in chemistry or solution concentration scenarios.

megacat8921
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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?
 
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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:

$$x+y=100$$

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:

$$0.6x+0.2y=50$$

or:

$$3x+y=250$$

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

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

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

MarkFL said:
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:

$$x+y=100$$

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:

$$0.6x+0.2y=50$$

or:

$$3x+y=250$$

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

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

Can you proceed?
 
megacat8921 said:
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
 

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