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Chapter 9: Problem 11

Give all solutions of each nonlinear system of equations, including those with nonreal complex components. $$\begin{aligned} &y=x^{2}-2 x+1\\\ &x-3 y=-1 \end{aligned}$$

Short Answer

Expert verified
The solutions are \( \left(\frac{1}{3}, \frac{4}{9}\right) \) and \( (2, 1) \).

Step by step solution

01

Solve the first equation for y

The first equation is given by \[ y = x^2 - 2x + 1 \]Rewriting this equation explicitly, we have:\[ y = (x-1)^2 \]
02

Substitute y in the second equation

Replace the value of \( y \) from the first equation into the second equation:\[ x - 3((x-1)^2) = -1 \]This substitution gives us the equation solely in terms of \( x \).
03

Simplify and solve for x

Simplify the equation:\[ x - 3(x^2 - 2x + 1) = -1 \]Expand the terms:\[ x - 3x^2 + 6x - 3 = -1 \]Combine like terms:\[ -3x^2 + 7x - 3 = -1 \]Move all terms to one side of the equation:\[ -3x^2 + 7x - 2 = 0 \]Solve the quadratic equation. By using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = -3 \), \( b = 7 \), and \( c = -2 \):\[ x = \frac{-7 \pm \sqrt{7^2 - 4(-3)(-2)}}{2(-3)} \]\[ x = \frac{-7 \pm \sqrt{49 - 24}}{-6} \]\[ x = \frac{-7 \pm \sqrt{25}}{-6} \]\[ x = \frac{-7 \pm 5}{-6} \]This gives us two solutions for \( x \):\[ x = \frac{-7 + 5}{-6} = \frac{-2}{-6} = \frac{1}{3} \]\[ x = \frac{-7 - 5}{-6} = \frac{-12}{-6} = 2 \]
04

Find corresponding y values

Substitute \( x = \frac{1}{3} \) and \( x = 2 \) back into the equation \( y = (x-1)^2 \):For \( x = \frac{1}{3} \):\[ y = \left(\frac{1}{3} - 1\right)^2 = \left(-\frac{2}{3}\right)^2 = \frac{4}{9} \]For \( x = 2 \):\[ y = (2-1)^2 = 1 \]
05

Write the final solutions

The solutions to the system of equations are given by the pairs:\( \left(\frac{1}{3}, \frac{4}{9}\right) \) and \( (2, 1) \).

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Key Concepts

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

quadratic equations
Understanding quadratic equations is crucial for solving nonlinear systems. A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants and \( a eq 0 \). In our nonlinear system, the first equation is quadratic: \( y = x^2 - 2x + 1 \). Quadratic equations have curved graphs called parabolas. These parabolas can open upwards or downwards, depending on the sign of the coefficient \( a \). When solving quadratic equations within a system, ensure to understand the quadratic nature to appropriately solve and interpret the results.
substitution method
One efficient way to solve a nonlinear system of equations, like ours, is through the substitution method. First, solve one equation for one variable. In our exercise, we solved the first equation for \( y \), giving us \( y = (x-1)^2 \). Substitute this expression into the other equation. This reduces our system to a single equation in one variable. It's like peeling an onion layer by layer: eliminating one variable allows us to focus on solving for the other. Be careful during substitution to simplify correctly to avoid mistakes.
complex solutions
Sometimes, solving a quadratic equation within a nonlinear system leads to complex solutions. Complex solutions appear when the discriminant (\( b^2 - 4ac \)) in the quadratic formula is negative. Complex numbers have a real part and an imaginary part (\( i \)), where \( i \) is defined as \( \sqrt{-1} \). In our exercise, we found real solutions, but always check the discriminant to determine if complex solutions arise. For example, if the discriminant were negative: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] would yield complex numbers.
quadratic formula
The quadratic formula is an essential tool for solving quadratic equations. It is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Let's break it down: \( b^2 - 4ac \) is called the discriminant. The discriminant tells us the nature of the roots. If positive, we get two real solutions; if zero, one real solution; if negative, two complex solutions. In our exercise, using the quadratic formula with \( a = -3 \), \( b = 7 \), and \( c = -2 \) led us to the solutions \( x = 2 \) and \( x = \frac{1}{3} \). Always plug in the respective \( a, b, \) and \( c \) values carefully to avoid computational errors.

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Most popular questions from this chapter

Given \(A=\left[\begin{array}{rr}4 & -2 \\ 3 & 1\end{array}\right], B=\left[\begin{array}{rr}5 & 1 \\ 0 & -2 \\ 3 & 7\end{array}\right],\) and \(C=\left[\begin{array}{rrr}-5 & 4 & 1 \\ 0 & 3 & 6\end{array}\right],\) find each product when possible. $$C B$$

Use the determinant theorems to find the value of each determinant. $$\left|\begin{array}{rrrr} 4 & 0 & 0 & 2 \\ -1 & 0 & 3 & 0 \\ 2 & 4 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}\right|$$

Solve each system by using the inverse of the coefficient matrix. $$\begin{array}{r} 6 x+9 y=3 \\ -8 x+3 y=6 \end{array}$$

$$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right], \quad B=\left[\begin{array}{ll} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}\right], \quad \text { and } \quad C=\left[\begin{array}{ll} c_{11} & c_{12} \\ c_{21} & c_{22} \end{array}\right] $$ where all the elements are real numbers. Use these matrices to show that each statement is true for \(2 \times 2\) matrices. \(A+B=B+A\) (commutative property)

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to determine the solution set. $$\begin{array}{r} x+y=4 \\ 2 x-y=2 \end{array}$$
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