Question

Solve this system of two equations with two unknowns using EES: $$ \begin{aligned} x^{3}-y^{2} &=10.5 \\ 3 x y+y &=4.6 \end{aligned} $$

Short answer

Based on the step-by-step solution above, provide a short answer to the problem: To solve the system of equations using the Elimination by Expansion System (EES) method, first rearrange both equations with one variable expressed in terms of the other. Then expand one of the equations to eliminate one of the variables and substitute the remaining equation to find the other variable's value. Finally, solve for the other variable, obtaining the values of x and y as the solutions to the system of equations. Note that a numerical method may be required to solve for x in this particular problem, and once x is found, simply substitute it back into the equation for y to find its value.

Step-by-step solution

Rearrange both equations

We can rearrange the first equation to express x in terms of y and the second equation to express y in terms of x: First equation: $$ x^3 = y^2 + 10.5 \Rightarrow x = \sqrt[3]{y^2 + 10.5} $$ Second equation: $$ 3xy + y = 4.6 \Rightarrow y(3x + 1) = 4.6 \Rightarrow y = \frac{4.6}{3x + 1} $$

Expand one of the equations

We will expand the first equation by squaring both sides in order to eliminate y: $$ (\sqrt[3]{y^2 + 10.5})^2 = y^{\frac{4}{3}} + 2y^{\frac{2}{3}}(10.5)^\frac{1}{3} + 10.5^{\frac{2}{3}} $$

Substitute the remaining equation

Now, we will substitute y in the expanded equation with the expression for y in terms of x from the second equation: $$ (\sqrt[3]{(\frac{4.6}{3x + 1})^2 + 10.5})^2 = (\frac{4.6}{3x + 1})^{\frac{4}{3}} + 2(\frac{4.6}{3x + 1})^{\frac{2}{3}}(10.5)^\frac{1}{3} + 10.5^{\frac{2}{3}} $$

Solve for x and y

At this point, we have a single equation with one unknown, x. To solve for x, use a numerical method such as the bisection method or Newton-Raphson method. Once the value of x is found, substitute it back into the expression for y in terms of x in the second equation: $$ y = \frac{4.6}{3x + 1} $$ Finally, we get the values of x and y, which are the solutions to the given system of equations.

Key concepts

Numerical Methods

Numerical methods are a type of mathematical tool used for finding approximate solutions to problems that may be difficult or impossible to solve analytically. These methods are particularly helpful in fields like engineering and physics, where exact solutions are not always feasible.

Key features of numerical methods include:

• Handling complex equations or systems of equations by providing approximate solutions.
• Flexibility in dealing with non-linear equations, which are often difficult to tackle analytically.
• Providing clarity and results even when equations don't have a neat, closed-form solution.

For example, in our exercise, numerical methods allow for solving the system of equations, despite their complexity.

Bisection Method

The bisection method is a simple and very effective numerical method for finding the roots of a continuous function. It works by repeatedly dividing an interval in half and selecting the subinterval where the function changes sign.

Here's how it works:

• Start with two points, say \(a\) and \(b\), such that the function at these points, \(f(a)\) and \(f(b)\), have opposite signs. This indicates that a root lies between \(a\) and \(b\).
• Compute the midpoint \(c = \frac{a+b}{2}\).
• Check the function value at \(c\). If \(f(c)\) is close enough to zero, \(c\) is the root. If not, then depending on the sign of \(f(c)\), choose the subinterval \([a, c]\) or \([c, b]\) for the next iteration.
• Repeat the process until the solution converges within the desired accuracy.

Although it might not be the fastest method for finding roots, its simplicity and reliability make it a handy option, especially when dealing with complicated equations.

Newton-Raphson Method

The Newton-Raphson method is another powerful technique for finding the roots of a real-valued function. Unlike the bisection method, it uses the function's derivative to find a better approximation of the root with each iteration.

The method steps are as follows:

• Start with an initial guess \(x_0\).
• Update the approximation using the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
• Repeat the process, updating the approximation until it converges to an accurate solution.

The Newton-Raphson method is faster than the bisection method, especially if the initial guess is close to the true root. However, it requires the derivative of the function, which can sometimes make it challenging to apply.

EES (Engineering Equation Solver)

EES, short for Engineering Equation Solver, is a software tool widely used by engineers for solving systems of non-linear algebraic or differential equations. It simplifies the process of managing and solving these equations by automating many of the complex calculations involved.

Important advantages of using EES include:

• Efficiently handles complex equations, reducing the time and effort required for solving them manually.
• Reduces errors, as it performs precise computational work that is less prone to human mistake.
• Provides many built-in functions to solve different kinds of thermodynamic and heat transfer problems commonly encountered in engineering.

In our exercise, utilizing EES would facilitate an easier resolution of the equations by automating the computational part, allowing focus on understanding the solution rather than on the manual calculation process.