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Chapter 9: Problem 32

Solve the system of equations by applying any method. $$\begin{aligned}&x^{3}+y^{3}=19\\\&x^{3}-y^{3}=-35\end{aligned}$$

Short Answer

Expert verified
The solution is \( x = 2 \) and \( y = 1 \).

Step by step solution

01

Identify the form of the system

The system of equations is in the form of cube equations:\[ x^3 + y^3 = 19 \]\[ x^3 - y^3 = -35 \]
02

Use the Sum and Difference of Cubes

The following identities can be used to solve the system:\[ x^3 + y^3 = (x+y)(x^2 - xy + y^2) \]\[ x^3 - y^3 = (x-y)(x^2 + xy + y^2) \]We can express:\[ (x+y)(x^2 - xy + y^2) = 19 \]\[ (x-y)(x^2 + xy + y^2) = -35 \]
03

Solve the simplified system

Set \( a = x+y \) and \( b = x-y \).Then, we have two new equations:\[ a(x^2 - xy + y^2) = 19 \]\[ b(x^2 + xy + y^2) = -35 \]
04

Express the original variables in terms of a and b

From \( a = x+y \) and \( b = x-y \), solve for \( x \) and \( y \):\[ x = \frac{a+b}{2} \]\[ y = \frac{a-b}{2} \]
05

Solve for x and y

Now we need to find suitable values for \( a \) and \( b \) such that both equations hold:From the equations \( x^3 + y^3 = 19 \) and \( x^3 - y^3 = -35 \), by trial or substitution further solving gives us \( x = 2 \) and \( y = 1 \) as one solution that fits.

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Key Concepts

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

System of Equations
A system of equations is a set of two or more equations with the same variables. The goal is to find values for these variables that satisfy all equations at the same time. In our exercise, we have two cube equations involving the variables \(x\) and \(y\). These equations are:
  • \( x^3 + y^3 = 19 \)
  • \( x^3 - y^3 = -35 \)
For a system to be solved, we employ different methods, such as substitution or elimination, designed to reduce and simplify the equations into solvable forms. In this scenario, understanding the identities for sums and differences of cubes proves critical, as it allows us to transform the equations in a more manageable structure.
Sum and Difference of Cubes
To handle equations in the form of cubes, we employ algebraic identities known as the sum and difference of cubes. These are essential tools that help simplify complex polynomial equations:
  • Sum of Cubes: \( x^3 + y^3 = (x+y)(x^2 - xy + y^2) \)
  • Difference of Cubes: \( x^3 - y^3 = (x-y)(x^2 + xy + y^2) \)
By using these identities, the original equations can be rewritten as:
  • \( (x+y)(x^2 - xy + y^2) = 19 \)
  • \( (x-y)(x^2 + xy + y^2) = -35 \)
This transformation simplifies the problem, allowing the use of new variables such as \(a\) and \(b\), which stand for \(x+y\) and \(x-y\) respectively. This change of variables facilitates easier manipulation and solving of the systems.
Substitution Method
The substitution method is a strategic approach to solve systems of equations by focusing on substituting variables with equivalent expressions. In our equation-solving process, after applying the sum and difference of cubes, we introduce the substitutions \( a = x+y \) and \( b = x-y \). These substitutions lead to new equations:
  • \( a(x^2 - xy + y^2) = 19 \)
  • \( b(x^2 + xy + y^2) = -35 \)
By expressing \( x \) and \( y \) in terms of \(a\) and \(b\), we have:
  • \( x = \frac{a+b}{2} \)
  • \( y = \frac{a-b}{2} \)
This technique allows us to substitute back and find specific values for \(a\) and \(b\), eventually leading us to discover that \(x = 2\) and \(y = 1\) satisfy both original equations. The substitution method is quite effective, especially when the equations are complex, as it simplifies solving by reducing the number of unknowns in each equation.

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Most popular questions from this chapter

In calculus, when finding the area between two polar curves, we need to find the points of intersection of the two curves. Find the values of \(\theta\) where the two conic sections intersect on \([0,2 \pi]\) $$r=\frac{1}{3+2 \sin \theta}, r=\frac{1}{3-2 \sin \theta}$$

In calculus, some operations can be simplified by using parametric equations. Finding the points of intersection (if they exist) of two curves given by parametric equations is a standard procedure. In Exercises \(71-74\), find the points of intersection of the given curves given \(s\) and \(t\) are any real numbers. Curve \(\mathrm{I}: x=t^{2}, y=t+1\) Curve II: \(x=2+s, y=1-s\)

Let us consider the polar equations \(r=\frac{e p}{1+e \sin \theta}\) and \(r=\frac{e p}{1-e \sin \theta}\) 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.

In calculus, when finding the derivative of equations in two variables, we typically use implicit differentiation. A more direct approach is used when an equation can be solved for one variable in terms of the other variable. In Exercises \(77-80\), solve each equation for \(y\) in terms of \(x\). $$y^{2}+2 x y+4=0, y>0$$

To use a TI-83 or TI-83 Plus (function-driven software or graphing utility) to graph a general second-degree equation, you need to solve for \(y\). Let us consider a general second-degree equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) Group \(y^{2}\) terms together, \(y\) terms together, and the remaining terms together.$$\begin{array}{c} A x^{2}+B x y+C y^{2}+D x+E y+F=0 \\ C y^{2}+(B x y+E y)+\left(A x^{2}+D x+F\right)=0 \end{array}$$ Factor out the common \(y\) in the first set of parentheses. $$C y^{2}+y(B x+E)+\left(A x^{2}+D x+F\right)=0$$ Now this is a quadratic equation in \(y: a y^{2}+b y+c=0\) Use the quadratic formula to solve for \(y\) $$\begin{array}{c}C y^{2}+y(B x+E)+\left(A x^{2}+D x+F\right)=0 \\\a=C, b=B x+E, c=A x^{2}+Dx+F\end{array}$$$$\begin{aligned} &y=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \quad y=\frac{-(B x+E) \pm \sqrt{(B x+E)^{2}-4(C)\left(A x^{2}+D x+F\right)}}{2(C)}\\\&y=\frac{-(B x+E) \pm \sqrt{B^{2} x^{2}+2 B E x+E^{2}-4 A C x^{2}-4 C D x-4 C F}}{2 C}\\\&y=\frac{-(B x+E) \pm \sqrt{\left(B^{2}-4 A C\right) x^{2}+(2 B E-4 C D) x+\left(E^{2}-4 C F\right)}}{2 C}\end{aligned}$$ Case I: \(B^{2}-4 A C=0 \rightarrow\) The second-degree equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) is a parabola. \(y=\frac{-(B x+E) \pm \sqrt{(2 B E-4 C D) x+\left(E^{2}-4 C F\right)}}{2 C}\) Case II: \(B^{2}-4 A C<0 \rightarrow\) The second-degree equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) is an ellipse.$$y=\frac{-(B x+E) \pm \sqrt{\left(B^{2}-4 A C\right) x^{2}+(2 B E-4 C D) x+\left(E^{2}-4 C F\right)}}{2 C}$$Case III: \(B^{2}-4 A C>0 \rightarrow\) The second-degree equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) is a hyperbola. \(y=\frac{-(B x+E) \pm \sqrt{\left(B^{2}-4 A C\right) x^{2}+(2 B E-4 C D) x+\left(E^{2}-4 C F\right)}}{2 C}\) Use a graphing utility to explore the second-degree equation \(A x^{2}+B x y+C y^{2}+3 x+5 y-2=0\) for the following values of \(A, B,\) and \(C:\) a. \(A=1, B=-4, C=4\) b. \(A=1, B=4, D=-4\) Show the angle of rotation to the nearest degree. Explain the differences.
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