Problem 9
Question
From \(x-y=0\) we have \(y=x .\) Substituting into \(3 x^{2}-4 y=0\) we obtain \(3 x^{2}-4 x=x(3 x-4)=0 .\) It follows that (0,0) and \((4 / 3,4 / 3)\) are the critical points of the system.
Step-by-Step Solution
Verified Answer
The critical points are \((0, 0)\) and \(\left( \frac{4}{3}, \frac{4}{3} \right)\)."
1Step 1: Understand the Problem
We are given two equations: \(x - y = 0\) and \(3x^2 - 4y = 0\). The task is to find the critical points by solving these equations simultaneously.
2Step 2: Express y in terms of x
From the equation \(x - y = 0\), we can express \(y\) in terms of \(x\):\[ y = x \].
3Step 3: Substitute y in the second equation
Substitute \(y = x\) into the second equation \(3x^2 - 4y = 0\):\[ 3x^2 - 4x = 0 \].
4Step 4: Factor the equation
Factor the equation \(3x^2 - 4x = 0\) by taking \(x\) common:\[ x(3x - 4) = 0 \].
5Step 5: Solve the factored equation
Set each factor of the equation \(x(3x - 4) = 0\) to zero:1. \(x = 0\)2. \(3x - 4 = 0\)Solving these gives \(x = 0\) and \(x = \frac{4}{3}\).
6Step 6: Determine y values
Using \(y = x\), find corresponding \(y\) values:- For \(x = 0\), \(y = 0\).- For \(x = \frac{4}{3}\), \(y = \frac{4}{3}\).
7Step 7: Identify Critical Points
The solutions \((x, y)\) give the critical points of the system:- \((0, 0)\)- \(\left( \frac{4}{3}, \frac{4}{3} \right)\).
Key Concepts
Simultaneous EquationsFactoring EquationsSystem of EquationsAlgebraic Substitution
Simultaneous Equations
Simultaneous equations are equations that are satisfied by the same values of variables. When solving these equations, we find a common solution that works for all equations in the system. This is essential in problems where two or more conditions must be true at the same time. In our exercise, the equations given are \(x - y = 0\) and \(3x^2 - 4y = 0\), and we aim to find the values of \(x\) and \(y\) that satisfy both equations simultaneously.
To tackle simultaneous equations, a typical approach is to either express one variable in terms of another or use both equations to eliminate one of the variables. This gives us a simpler equation to solve.
To tackle simultaneous equations, a typical approach is to either express one variable in terms of another or use both equations to eliminate one of the variables. This gives us a simpler equation to solve.
- Expressing a variable: Convert one equation to express a variable in terms of another, like transforming \(x - y = 0\) to \(y = x\).
- Substituting the expression into the other equation, simplifying the problem to a single equation, which completes the process of solving simultaneously.
Factoring Equations
Factoring equations involves breaking down complex expressions into simpler, multiplied factors. This technique helps us solve polynomial equations by setting each factor equal to zero.
In our exercise, after substituting \(y = x\) into \(3x^2 - 4y = 0\), we simplify it to \(3x^2 - 4x = 0\). The next step is factoring, where we take \(x\) as a common factor:
In our exercise, after substituting \(y = x\) into \(3x^2 - 4y = 0\), we simplify it to \(3x^2 - 4x = 0\). The next step is factoring, where we take \(x\) as a common factor:
- We factor out \(x\) to get \(x(3x - 4) = 0\).
- This results in two separate simple equations: \(x = 0\) and \(3x - 4 = 0\).
- Factoring simplifies the solving process as we set each factor to zero, allowing us to find the solutions for \(x\).
System of Equations
A system of equations is a collection of two or more equations with the same set of variables. Solving a system of equations means finding a set of values for the variables that satisfies all the equations simultaneously.
In our problem, the system includes two equations: \(x - y = 0\) and \(3x^2 - 4y = 0\). The goal is to identify the points \((x, y)\) where these equations intersect.
In our problem, the system includes two equations: \(x - y = 0\) and \(3x^2 - 4y = 0\). The goal is to identify the points \((x, y)\) where these equations intersect.
- First, determine the relationship between the variables using one equation (e.g., \(y = x\) from \(x - y = 0\)).
- Next, apply this relationship to another equation, converting the system from two equations in two variables to one equation in a single variable.
Algebraic Substitution
Algebraic substitution is a method used to simplify and solve systems of equations. It involves solving one of the equations for one variable and then substituting that expression into another equation.
In our example, we start by solving the simplest equation, \(x - y = 0\), to get \(y = x\). We then substitute \(y = x\) into the second equation, \(3x^2 - 4y = 0\), to transform it into a single variable equation, \(3x^2 - 4x = 0\).
In our example, we start by solving the simplest equation, \(x - y = 0\), to get \(y = x\). We then substitute \(y = x\) into the second equation, \(3x^2 - 4y = 0\), to transform it into a single variable equation, \(3x^2 - 4x = 0\).
- The substitution method is particularly useful when one equation is simple and can be rearranged easily.
- By reducing the system from two variables to one, it simplifies the algebra required to solve the system.
Other exercises in this chapter
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 Problem 10
(a) Solving $$\begin{array}{l} x(-0.1+0.02 y)=0 \\ y(0.2-0.025 x)=0 \end{array}$$ in the first quadrant we obtain the critical point \((8,5) .\) The graphs are
View solution Problem 11
Solving $$\begin{array}{l} x(20-0.4 x-0.3 y)=0 \\ y(10-0.1 y-0.3 x)=0 \end{array}$$ we see that critical points are \((0,0),(0,100),(50,0),\) and \((20,40) .\)
View solution