Problem 7
Question
From \(x+x y=0\) we have \(x(1+y)=0 .\) Therefore \(x=0\) or \(y=-1 .\) If \(x=0,\) then, substituting into \(-y-x y=0\) we obtain \(y=0 .\) Likewise, if \(y=-1,1+x=0\) or \(x=-1 .\) We can conclude that (0,0) and (-1,-1) are critical points of the system.
Step-by-Step Solution
Verified Answer
The critical points are (0,0) and (-1,-1).
1Step 1: Understanding the Given Equation
We start with the equation \(x + xy = 0\). Our goal is to find the values of \(x\) and \(y\) that satisfy this equation.
2Step 2: Factoring the Equation
Factor the equation as follows: \(x(1+y) = 0\). From this, either \(x = 0\) or \(1 + y = 0\).
3Step 3: Solving for \(x = 0\)
If \(x = 0\), substitute \(x = 0\) in the second equation \(-y - xy = 0\). Since \(x = 0\), the equation simplifies to \(-y = 0\). Thus, \(y = 0\).
4Step 4: Solving for \(y = -1\)
If \(1 + y = 0\), then \(y = -1\). Substitute \(y = -1\) in the equation \(1 + x = 0\). This simplifies to \(x = -1\).
5Step 5: Identifying Critical Points
The solutions \((x, y) = (0, 0)\) and \((x, y) = (-1, -1)\) are found at the points where the original system of equations is satisfied. These points are the critical points.
Key Concepts
Factoring EquationsSystem of EquationsSubstitution Method
Factoring Equations
Factoring equations is an essential technique in algebra. It allows us to simplify and solve equations by breaking them down into more manageable components. In the equation given in the exercise, we start with \( x + xy = 0 \). The goal of factoring here is to identify common factors that can help us rewrite the equation into a simpler form. By factoring, we can express the original equation as \( x(1 + y) = 0 \).
This factored form implies that at least one of these factors must be zero. In other words, either \( x = 0 \) or \( 1 + y = 0 \). Breaking down an equation into factors simplifies the process of finding solutions, as it reduces the task into smaller parts that are easier to solve. Factoring is a key step not only in solving algebraic equations but also in identifying critical points within systems of equations.
This factored form implies that at least one of these factors must be zero. In other words, either \( x = 0 \) or \( 1 + y = 0 \). Breaking down an equation into factors simplifies the process of finding solutions, as it reduces the task into smaller parts that are easier to solve. Factoring is a key step not only in solving algebraic equations but also in identifying critical points within systems of equations.
System of Equations
A system of equations consists of multiple equations that need to be solved simultaneously. In this exercise, we are dealing with two equations: \( x + xy = 0 \) and \( -y - xy = 0 \). The solutions to this system are the values of \( x \) and \( y \) that satisfy both equations at the same time.
The point of finding a solution to a system of equations is to determine points where the equations intersect, which in this case are called critical points. When solving systems of equations, using methods such as substitution or elimination can help isolate and solve for each variable, allowing us to understand the relationship between them.
The point of finding a solution to a system of equations is to determine points where the equations intersect, which in this case are called critical points. When solving systems of equations, using methods such as substitution or elimination can help isolate and solve for each variable, allowing us to understand the relationship between them.
Substitution Method
The substitution method is a powerful technique used to solve systems of equations. It involves solving one equation for one variable and then substituting this expression into the other equation. This helps reduce the complexity of the system by effectively working with just one variable at a time.
For example, in the exercise given, once we determine that \( x = 0 \) from \( x(1 + y) = 0 \), we can substitute \( x = 0 \) into the second equation \( -y - xy = 0 \) to simplify the problem to \( -y = 0 \). This immediately gives us \( y = 0 \). Similarly, if we find \( y = -1 \) from \( 1 + y = 0 \), this information can be substituted into another related equation to find \( x = -1 \).
Substitution is especially useful for solving systems where substitution simplifies complicated equations into straightforward arithmetic, returning straightforward solutions for each variable.
For example, in the exercise given, once we determine that \( x = 0 \) from \( x(1 + y) = 0 \), we can substitute \( x = 0 \) into the second equation \( -y - xy = 0 \) to simplify the problem to \( -y = 0 \). This immediately gives us \( y = 0 \). Similarly, if we find \( y = -1 \) from \( 1 + y = 0 \), this information can be substituted into another related equation to find \( x = -1 \).
Substitution is especially useful for solving systems where substitution simplifies complicated equations into straightforward arithmetic, returning straightforward solutions for each variable.
Other exercises in this chapter
Problem 5
The corresponding plane autonomous system is \\[ x^{\prime}=y, \quad y^{\prime}=-x+\epsilon x^{3} \\] If \((x, y)\) is a critical point, \(y=0\) and \(-x+\epsil
View solution Problem 6
The corresponding plane autonomous system is \\[ x^{\prime}=y, \quad y^{\prime}=-x+\epsilon x|x| \\] If \((x, y)\) is a critical point, \(y=0\) and \(-x+\epsilo
View solution Problem 8
The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=-x+\left(\frac{1}{2}+3 y^{2}\right) y-x^{2}$$ and so \(\frac{\partial P}{\partial
View solution Problem 8
From \(y^{2}-x=0\) we have \(x=y^{2} .\) Substituting into \(x^{2}-y=0,\) we obtain \(y^{4}-y=0\) or \(y\left(y^{3}-1\right)=0 .\) It follows that \(y=0,1\) and
View solution