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Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{l} {y=x^{2}-4} \\ {y=x^{2}-4 x} \end{array}\right. $$

Short Answer

Expert verified
The real solution is \((1, -3)\).

Step by step solution

01

Equate the Equations

Given the system of equations, \(y = x^2 - 4\) and \(y = x^2 - 4x\), to find the points where both equations are equal, set the right-hand sides of the equations equal to each other: \[x^2 - 4 = x^2 - 4x.\]
02

Simplify the Equation

Subtract \(x^2\) from both sides: \[x^2 - 4 - x^2 = x^2 - 4x - x^2.\] This simplifies to: \[-4 = -4x.\]
03

Solve for x

Divide both sides by -4 to isolate \(x\): \[x = 1.\]
04

Substitute x into One of the Original Equations to Find y

Substitute \(x = 1\) back into the first equation: \[y = x^2 - 4 = 1^2 - 4 = 1 - 4 = -3.\] So, the solution is \((x, y) = (1, -3).\)

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Key Concepts

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

Solving Nonlinear Equations
When it comes to solving nonlinear systems of equations, the process typically involves finding a solution that satisfies all the given equations simultaneously. Nonlinear equations differ from linear ones because they can contain variables that are raised to a power other than one.
In many problems, the goal is to equate two or more nonlinear equations to find a common solution. This involves understanding the relationships described by the equations and manipulating them to isolate the unknowns.

The nonlinear equations given in the exercise can be visualized as curves on a graph. The intersection point(s) of these curves represent the solution(s) to the system. So effectively, solving a nonlinear system like the one given means finding where these curves meet.
  • This involves equating the equations.
  • Simplifying to solve for one variable.
  • Substituting back to find other variable values.
Nonlinear systems can often be more complex than linear ones due to the nature of the curves involved, leading sometimes to multiple solutions or no real solution at all.
Algebraic Manipulation
Algebraic manipulation is a crucial tool in solving equations, especially when dealing with systems of nonlinear equations. It involves performing operations that simplify the equations, making it easier to isolate and solve for the unknown variables.
For instance, in the provided exercise, equating the two given equations allowed for simplifying and removing the quadratic term, leading to a simpler, linear equation: \(-4 = -4x\).

Let's break it down further:
  • Start by setting the equations equal to each other when both describe the same dependent variable, in this case, \(y\).
  • Perform algebraic operations such as subtraction on both sides to eliminate terms.
  • Isolate the variable by dividing or multiplying both sides as needed.
This systematic approach helps us solve for one variable at a time, which is key to deciphering nonlinear equation systems. Such methods form the bedrock upon which more complex algebraic techniques are built.
System of Equations
A system of equations involves multiple equations working in tandem to find values for their variables that satisfy them collectively. In the context of the exercise, it's a system of nonlinear equations.
Here, you have two equations, both defining the same variable, \(y\). Each equation represents a different curved graph on a Cartesian plane. The solution to this system can be found at the intersection point of these two curves.

Here's how to tackle systems of equations:
  • Write down all equations in the system clearly.
  • Identify a common variable or term to eliminate in order to simplify.
  • Equate or rearrange equations to work towards a solution.
Solving systems of equations requires logical reasoning and precise algebraic manipulation to extract the values of unknowns.
Quadratic Equations
Quadratic equations are a specific type of nonlinear equation where the highest degree of any variable is two. They are typically written in the form \(ax^2 + bx + c = 0\). In our system, each equation is quadratic, providing richness to the problem because their graphs are parabolas.
Quadratic equations like those in the exercise create parabolic paths due to the squared term. Solving them within systems involves:
  • Recognizing each equation's standard quadratic form.
  • Using algebraic manipulation to reduce them, if possible.
  • Finding graphical intersections, reflecting real number solutions.
This exercise highlights the elegance and complexity of quadratic equations within nonlinear systems. Quadratics frequently appear in both practical problems and theoretical studies, making understanding them essential in mathematics deep learning.

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Most popular questions from this chapter

Rationalize each denominator and simplify if possible. $$\frac{1}{\sqrt{3}}$$

For the exercises below, see the Concept Check in this section. How many real solutions are possible for a system of equations whose graphs are an ellipse and a line? Draw diagrams to illustrate each possibility.

Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{r} {x^{2}+y^{2}=1} \\ {x^{2}+(y+3)^{2}=4} \end{array}\right. $$

The orbits of stars, planets, comets, asteroids, and satellites all have the shape of one of the conic sections. Astronomers use a measure called eccentricity to describe the shape and elongation of an orbital path. For the circle and ellipse, eccentricity e is calculated with the formula \(e=\frac{c}{d},\) where \(c^{2}=\left|a^{2}-b^{2}\right|\) and \(d\) is the larger value of a or b. For a hyperbola, eccentricity e is calculated with the formula \(e=\frac{c}{d},\) where \(c^{2}=a^{2}+b^{2}\) and the value of \(d\) is equal to a if the hyperbola has \(x\) -intercepts or equal to b if the hyperbola has \(y\) -intercepts. A. \(\frac{x^{2}}{36}-\frac{y^{2}}{13}=1\) B. \(\frac{x^{2}}{4}+\frac{y^{2}}{4}=1\) C. \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) D. \(\frac{y^{2}}{25}-\frac{x^{2}}{39}=1\) G. \(\frac{x^{2}}{16}-\frac{y^{2}}{65}=1\) E. \(\frac{x^{2}}{17}+\frac{y^{2}}{81}=1\) F. \(\frac{x^{2}}{36}+\frac{y^{2}}{36}=1\) H. \(\frac{x^{2}}{144}+\frac{y^{2}}{140}=1\) What do you notice about the values of \(e\) for the equations you identified as hyperbolas?

Find each function value if \(f(x)=3 x^{2}-2 .\) See Section 3.2 $$ f(-1) $$

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