/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Solve the system of equations. ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 4

Solve the system of equations. $$\begin{array}{c} 4 p_{1}-2 p_{2}=0 \\ p_{1}+p_{2}=1 \end{array}$$ For larger matrices, you can edit the Sage cell by adding additional terms in each row, and adding more rows. For example, you can replace [3,-5,1],[1,1,3] with [4,8,-4,4],[3,8,5,-11],[-2,1,12,-17].

Short Answer

Expert verified
\( p_1 = \frac{1}{3}, \; p_2 = \frac{2}{3} \)

Step by step solution

01

Express One Variable in Terms of the Other

First, we'll express one variable in terms of the other using the second equation. We have:\[ p_1 + p_2 = 1 \]Let's solve for \( p_1 \):\[ p_1 = 1 - p_2 \]
02

Substitute the Expression

Substitute \( p_1 = 1 - p_2 \) into the first equation:\[ 4(1 - p_2) - 2p_2 = 0 \]Expand the equation:\[ 4 - 4p_2 - 2p_2 = 0 \]
03

Simplify the Equation

Combine like terms:\[ 4 - 6p_2 = 0 \]Now, solve for \( p_2 \) by subtracting 4 from both sides:\[ -6p_2 = -4 \]
04

Solve for \\( p_2 \\)

Divide both sides by -6 to solve for \( p_2 \):\[ p_2 = \frac{-4}{-6} = \frac{2}{3} \]
05

Solve for \\( p_1 \\)

Using the expression \( p_1 = 1 - p_2 \), substitute \( p_2 = \frac{2}{3} \):\[ p_1 = 1 - \frac{2}{3} = \frac{1}{3} \]
06

Verify the Solution

Substitute \( p_1 = \frac{1}{3} \) and \( p_2 = \frac{2}{3} \) back into the original equations to check:1. \( 4 \left( \frac{1}{3} \right) - 2 \left( \frac{2}{3} \right) = \frac{4}{3} - \frac{4}{3} = 0 \)2. \( \frac{1}{3} + \frac{2}{3} = 1 \)Both equations are satisfied, confirming the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation Solving
Solving a linear system of equations involves finding the values of the variables that satisfy all the given equations simultaneously. In our example, you will find values for \(p_1\) and \(p_2\) such that both equations are true at the same time. Think of these equations as lines on a graph. The solution to the system is the point where the lines intersect. This point represents the values of the variables that satisfy all equations in the system. Often, solving such systems is done by either graphing, substitution, or elimination methods. Each of these techniques serves a purpose and understanding how they work is essential for solving linear equations efficiently.
The key to solving equations is to perform operations that do not affect the equality of the equations. By transforming these equations step-by-step, we aim to isolate one of the variables, making it easier to find a solution. In our exercise, we first solved for \(p_1\) in terms of \(p_2\), which helped simplify the process. This technique allows us to gradually work towards finding a singular solution that satisfies both equations.
Substitution Method
The substitution method is a straightforward technique for solving linear systems, especially when you can easily solve one of the equations for one variable. In our problem, we started by isolating \(p_1\) from the second equation:
  • Start by expressing \(p_1\) in terms of \(p_2\) from the equation \( p_1 + p_2 = 1 \), resulting in \( p_1 = 1 - p_2 \).
  • Substitute this expression into the first equation. This step is crucial as it reduces the number of variables in the equation, simplifying it to a single variable equation: \(4(1 - p_2) - 2p_2 = 0 \).
Once substituted, manage the equation by expanding and combining like terms to solve for the remaining variable. This reduction allows us to determine \(p_2\) first, and then \(p_1\) can subsequently be found using the expression derived earlier. This method leverages the idea of breaking down a complex system into simpler parts, making it easier to solve.
Verification of Solutions
Verification is an essential step in solving linear equations as it ensures that the obtained solution truly satisfies the original system of equations. Once both \(p_1\) and \(p_2\) are found, substitute these values back into the original equations.
  • For our solution, verify by substituting \(p_1 = \frac{1}{3}\) and \(p_2 = \frac{2}{3}\) into the equations: \(4p_1 - 2p_2 = 0 \) and \(p_1 + p_2 = 1 \).
  • Each substitution confirms whether the equations hold true with the calculated values. This serves as a double-check, reflecting the correctness or needs for revision in the calculations.
Verification provides assurance that no mistakes occurred, and it guarantees that the solution aligns with the original mathematical model presented by the equations. Even after finding values that satisfy the manipulated forms of the equations, checking against the original equations is necessary for ultimate confirmation.

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Most popular questions from this chapter

Expected value of conjecture. Suppose both players play your ideal strategy in the Matching Pennies game, what should the expected value of the game be? We could use our previous graphical method to determine the expected value of the game (you might quickly try this just to verify your prediction). However, as we have noted, a major drawback of the graphical solution is that if our players have 3 (or more) options, then we would need to graph an equation in 3 (or more!) variables; which, I hope you agree, we don't want to do. Although we will continue to focus on \(2 \times 2\) games, we will develop a new method which can more easily be used to solve to larger games. We will need a little notation. Let \(\begin{aligned} P_{1}(H) &=\text { the probability that Player } 1 \text { plays } \mathrm{H} \\ P_{1}(T) &=\text { the probability that Player } 1 \text { plays } \mathrm{T} ; \\ P_{2}(H) &=\text { the probability that Player } 2 \text { plays } \mathrm{H} ; \\ P_{2}(T) &=\text { the probability that Player } 2 \text { plays } \mathrm{T} \end{aligned}\) Also, we will let \(E_{1}(H)\) be the expected value for Player 1 playing pure strategy H against a given strategy for Player 2. Similarly, \(E_{2}(H)\) will be Player 2's expected value for playing pure strategy \(\mathrm{H}\).

In a repeated game, if a player always plays the same row (or column), we say that she is playing a pure strategy. \(\Delta\) For example, if Player 1 always plays Row \(\mathrm{A},\) we say she is playing pure strategy \(A\).

A \(2 \times 2\) Repeated Game. Consider the following zero-sum game $$\left[\begin{array}{cc} 1 & 0 \\ -1 & 2 \end{array}\right]$$ Does this game have an equilibrium pair? Play this game with an opponent 10 times. Tally your wins and losses. Describe how you chose which strategy to play. Describe how your opponent chose which strategy to play. When playing the game several times, does it make sense for either player to play the same strategy all the time? Why or why not? Although we use the term "strategy" to mean which row (or column) a player chooses to play, we will also refer to how a player plays a repeated game as the player's strategy. In order to avoid confusion, in repeated games we will define some specific strategies.

First sketch the axes. (Recall, the \(x\) -axis only goes from 0 to \(1 .)\)

Consider the system of equations $$\begin{array}{cl} 2 x+y-4 z & =10 \\ 3 x+5 z & =-5 \\ y+2 z & =7 \end{array}$$ a Set up the augmented matrix for this system. b Use row reduction to find the solution.
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