/*! 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 1 An example of a system of two li... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 1

An example of a system of two lines: $$ \begin{array}{l} y=\frac{3}{5} x-\frac{1}{5} \\ y=-x+3 \end{array} $$

Short Answer

Expert verified
The lines intersect at the point (2, 1).

Step by step solution

01

Set the Equations Equal

To find the intersection of two lines, we need to set their equations equal because at the intersection point, the y-values of both equations are the same. Hence, set \(\frac{3}{5}x - \frac{1}{5} = -x + 3\).
02

Eliminate Fractions

To eliminate the fractions, multiply every term by 5 (the common denominator). This gives us \(3x - 1 = -5x + 15\).
03

Rearrange Terms

Next, rearrange the terms to get all x's on one side and constants on the other: \(3x + 5x = 15 + 1\).
04

Simplify the Equation

Simplify both sides to get \(8x = 16\).
05

Solve for x

Divide both sides by 8 to solve for x: \(x = 2\).
06

Substitute x into One Equation

Substitute \(x = 2\) into one of the original equations, for example \(y = -x + 3\). This gives \(y = -2 + 3\).
07

Solve for y

Solve for \(y\) to find \(y = 1\).
08

Write the Solution

The solution to the system, or the point of intersection, is \((x, y) = (2, 1)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Intersection Point
When discussing systems of equations, finding the intersection point of two lines is crucial. This point is where both line equations have the same value for both the x and y coordinates. Imagine a busy road intersection; it's the same idea鈥攁 spot where paths meet. To find this point, we set the equations equal, because at this point, their corresponding y-values are the same. For example, with our given lines, we started by equating the two expressions for y:
  • First equation: \( y = \frac{3}{5} x - \frac{1}{5} \)
  • Second equation: \( y = -x + 3 \)
We aim for a single equation in one variable (typically x or y) as this will reveal the specific value at which the lines cross.
Eliminate Fractions
Fractions can complicate solving equations, so it's often best to eliminate them early on. The trick is to multiply every term by a common multiple (usually the smallest common denominator) to shift to whole numbers, making equations easier to handle.
In our example, the common denominator in the first equation is 5. Multiplying everything by 5 frees us from fractions:
  • Original: \( \frac{3}{5}x - \frac{1}{5} = -x + 3 \)
  • After multiplying by 5: \( 3x - 1 = -5x + 15 \)
Remember, this step doesn't change the equation's inherent relationships; it simply scales everything. Now, it's simpler to work through without fractions distracting us.
Solve for Variables
With fractions eliminated, solving for variables is straightforward. Our goal is to isolate a single variable, typically x or y, on one side of the equation. In the reorganized example equation:
  • Combine like terms to find \( 8x = 16 \)
  • Divide both sides by 8 to get \( x = 2 \)
Next, substitute this value back into one of the original equations to solve for the other variable. Here, with \( x = 2 \), substitute into \( y = -x + 3 \):
  • You get \( y = -2 + 3 \)
  • Solve to find \( y = 1 \)
Thus, both variables are determined, and you have reached your solution.
Graphing Linear Equations
Graphing linear equations is like drawing a map for solutions. By graphing, you visually identify where curves intersect, representing system solutions. With equations as straight lines, the intersection point is determined by where they cross.
For our system:
  • Graph \( y = \frac{3}{5}x - \frac{1}{5} \); it slopes upward, starting just below the y-axis.
  • Graph \( y = -x + 3 \); it slopes downward from up on the y-axis.
The lines intersect at \((2, 1)\), aligning with our algebraic finding. This visual tool confirms that understanding the algebraic result matches the graph's display, reinforcing the solution's validity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve the system of equations. $$ \begin{array}{c} 2 x-2 y=6 \\ x+3 y=7 \end{array} $$

Consider the system of equations $$\begin{array}{cc} 3 x+2 y-w-v & =0 \\ 2 x-y+3 z+w+5 v & =0 \\ x+2 y+6 z-w & =0 \\ -y+z-3 w+v & =0 \\ x+y+z+w+v & =1 \end{array}$$ a Set up the augmented matrix for this system. b Use row reduction to find the solution. Now we are ready to apply everything we have learned about solving repeated zero-sum games to a much more challenging game in the next section!

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 \(\mathrm{H}\).

How might we determine a "winner" for Undercut after playing several times? Most likely, you said that someone will win the game if they have the most points. In fact, we probably don't care if the final score is \(10-12\) or \(110-112 .\) In either case, Player 2 wins. Since we will play this game several times, we do care about the point difference. For example, a score of \(5-1\) would be better for Player 1 than 5-3. So let's think about the game in terms of the point difference between the players in a given game. This is called the net gain. For example, with score of 5 - 1 , Player 1 would have a net gain of 4 .

We begin with a deck of cards which has \(50 \%\) aces (A) and \(50 \%\) kings (K). Aces rank higher than kings. Player 1 is dealt one card, face down. Player 1 can look at the card, but does not show the card to Player 2. Player 1 then says "ace" or "king" depending on what his card is. Player 1 can either tell the truth and say what the card is \((\mathrm{T}),\) or he can bluff and say that he has a higher ranking card (B). Note that if Player 1 has an ace, he must tell the truth since there are no higher ranking cards. However, if he is dealt a king, he can bluff, by saying he has an ace. If Player 1 says "king" the game ends and both players break even. If Player 1 says "ace" then Player 2 can either call (C) or fold (F). If Player 2 folds, then Player 1 wins $$\$ 0.50$$If Player 2 calls and Player 1 does not have an ace, then Player 2 wins $$\$ 1 .$$ If Player 2 calls and Player 1 does have an ace, then Player 1 wins $$\$ 1 . $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.