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Chapter 9: Problem 21

Solve the system of equations by applying the substitution method. $$\begin{array}{r}2 x-y=3 \\\x^{2}+y^{2}-2 x+6 y=-9\end{array}$$

Short Answer

Expert verified
The solutions are \((0, -3)\) and \((\frac{2}{5}, -\frac{11}{5})\).

Step by step solution

01

Solve the First Equation for y

The first equation is \(2x - y = 3\). Let's isolate \(y\) by subtracting \(2x\) from both sides to get \( -y = 3 - 2x \). Then, multiply by \(-1\) to obtain \( y = 2x - 3 \).
02

Substitute y in the Second Equation

After finding \(y = 2x - 3\), substitute this expression into the second equation \(x^2 + y^2 - 2x + 6y = -9\). This yields \(x^2 + (2x - 3)^2 - 2x + 6(2x - 3) = -9\).
03

Simplify and Solve the Quadratic Equation

First, expand the expression \((2x - 3)^2\) to get \(4x^2 - 12x + 9\). Substitute back into the equation to have \(x^2 + 4x^2 - 12x + 9 - 2x + 12x - 18 = -9\). Simplifying, we combine like terms: \(5x^2 - 2x - 9 = -9\).
04

Solve for x

Add 9 to both sides of the equation \(5x^2 - 2x - 9 = -9\) to get \(5x^2 - 2x = 0\). Factor the left side by taking \(x\) as a common factor: \(x(5x - 2) = 0\). This gives the solutions \(x = 0\) or \(5x - 2 = 0\). Solving for \(x\) in the second equation gives \(x = \frac{2}{5}\).
05

Solve for y using x-values

We previously found \(y = 2x - 3\). For \(x = 0\), substitute into \(y = 2(0) - 3 = -3\). For \(x = \frac{2}{5}\), substitute into \(y = 2(\frac{2}{5}) - 3 = \frac{4}{5} - 3 = -\frac{11}{5}\). Thus, the solutions for \(y\) are \(y = -3\) and \(y = -\frac{11}{5}\).
06

List the Solution Pairs

The solution pairs for this system of equations are \((x, y) = (0, -3)\) and \((x, y) = \left(\frac{2}{5}, -\frac{11}{5}\right)\).

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Key Concepts

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

Introduction to System of Equations
In mathematics, a system of equations is a set of two or more equations with the same variables. Solving a system of equations means finding the values for the variables that satisfy all the given equations simultaneously. Systems are often classified by the number of equations and unknowns:
  • Linear Systems: All equations are linear.
  • Non-linear Systems: One or more equations are non-linear.
To solve these systems, several methods exist, such as:
  • Substitution Method: Solve one equation for one variable and substitute this into another equation.
  • Elimination Method: Add or subtract equations to eliminate one variable.
  • Graphical Method: Graph the equations to find intersection points.
This exercise involves a combination of linear and quadratic equations鈥攔epresenting a non-linear system.
Understanding and Solving Quadratic Equations
A quadratic equation is a type of polynomial equation where the highest exponent of the variable is 2. It generally takes the form of \( ax^2 + bx + c = 0 \).
Key features of quadratic equations include:
  • They form parabolas when graphed.
  • They have a highest exponent of 2 (degree 2).
Quadratic equations can be solved using different methods:
  • Factoring: Express the quadratic as a product of its linear factors, if possible.
  • Completing the Square: Reorganize the equation to make one side a perfect square trinomial.
  • Quadratic Formula: Use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
In the context of the exercise, we arrive at a quadratic equation, \(5x^2 - 2x = 0\), which can be solved by factoring.
Solving for Variables Using Substitution Method
The substitution method is a powerful technique for solving systems of equations. It involves replacing one variable in one equation with an equivalent expression derived from another equation. This helps in reducing the system to a single equation with one variable.
Here's how it's applied:
  • Solve one equation for one variable.
  • Substitute this expression into the other equation(s).
  • Solve for the remaining variable.
  • Back-substitute to find the other variable values.
In this problem, we solve the first equation, \(2x - y = 3\), for \(y\) to get \(y = 2x - 3\). Then, substitute \(y\) in the second equation \(x^2 + y^2 - 2x + 6y = -9\). The substitution reduces it to a quadratic equation in terms of \(x\). Finally, solving this gives the solutions for \(x\), which allows calculating \(y\) values by back-substitution.

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Most popular questions from this chapter

Let us consider the polar equations \(r=\frac{t p}{1+e \cos \theta}\) and \(r=\frac{e p}{1-e \cos \theta}\) with eccentricity \(e=1 .\) With a graphing utility, explore the equations with \(p=1,2,\) and \(6 .\) Describe the behavior of the graphs as \(p \rightarrow \infty\) and also the difference between the two equations.

Solve the system of equations by applying any method. $$\begin{aligned}&2 x^{2}+4 y^{4}=-2\\\&6 x^{2}+3 y^{4}=-1\end{aligned}$$

Graph the second-degree equation. (Hint: Transform the equation into an equation that contains no \(x y\) -term.) $$3 x^{2}+2 \sqrt{3} x y+y^{2}+2 x-2 \sqrt{3} y-12=0$$

Let us consider the polar equations \(r=\frac{e p}{1+e \cos \theta}\) and \(r=\frac{e p}{1-e \cos \theta}\) with \(p=1 .\) With a graphing utility, explore the equations with \(e=0.001,0.5,0.9,\) and 0.99 Describe the behavior of the graphs as \(e \rightarrow 1\) and also the difference between the two equations. Be sure to set the window parameters properly.

Two boxes are constructed to contain the same volume. In the first box, the width is \(16 \mathrm{cm}\) larger than the depth and the length is five times the depth. In the second box, both the length and width are \(4 \mathrm{cm}\) shorter and the depth is \(25 \%\) larger than in the first box. Find the dimensions of the second box.
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