Problem 45 - Algebra and Trigonometry

Question

Let \(x\) represent one number and let \(y\) represent the other number. Use the given conditions to write a system of nonlinear equations. Solve the system and find the numbers. The difference between the squares of two numbers is \(3 .\) Twice the square of the first number increased by the square of the second number is \(9 .\) Find the numbers.

Step-by-Step Solution

Answer

The solutions to the system are (x, y): (2, 1), (-2, 1), (2, -1), (-2, -1)

Step 1: Translate Words into Mathematical Expressions

Transform the given conditions into mathematical expressions: \n 1) The difference between the squares of two numbers is 3, translates to \(x^2 - y^2 = 3\) \n 2) Twice the square of the first number increased by the square of the second number is 9, transforms into \(2x^2 + y^2 = 9\)

Step 2: Solve the First Equation for One Variable

Solving the first equation \(x^2 - y^2 = 3\) for \(x^2\) will make it easier to substitute into the second equation. It becomes \(x^2 = y^2 + 3\)

Step 3: Substitute the Expression of \(x^2\) into the Second Equation

Replace \(x^2\) in the second equation \(2x^2 + y^2 = 9\) with the expression derived from the first equation which is \(y^2 + 3\). The equation becomes \(2(y^2 + 3) + y^2 = 9\) which simplifies further to \(3y^2 + 6 - 9 = 0\) and finally becomes \(3y^2 - 3 = 0\)

Step 4: Solve for the Other Variable

Solving for \(y\) in the equation \(3y^2 - 3 = 0\) gives us two possible values: \(1\) and \(-1\) when simplified. Hence, \(y\) could be either \(1\) or \(-1\)

Step 5: Solve for the First Variable Using the Derived Values

Substitute the derived values of \(y\) into the earlier equation \(x^2 = y^2 + 3\). When \(y = 1\), the equation becomes \(x^2 = 1^2 + 3\) which yields \(x = \pm 2\). When \(y = -1\), the equation becomes \(x^2 = (-1)^2 + 3\) which also yields \(x = \pm 2\)

Key Concepts

Solving Nonlinear Equations

When dealing with nonlinear equations, we're working with mathematical expressions where variables are raised to a power other than one. This makes the equation nonlinear. In this exercise, we have two nonlinear equations defined by squares of variables. Solving these equations involves manipulating them to find values of the unknowns.

Here's how you can approach such equations:

• First, express one variable in terms of the other using an equation. This helps in eliminating one unknown.
• Substitute this expression into the other equation. This reduces the problem into solving a single nonlinear equation.
• Solve for the remaining variable, and then backtrack to find the other variable using the expression you derived earlier.

Keep in mind that the solutions to nonlinear equations can sometimes yield more than one set of answers, which should be considered carefully based on the context of the problem.

System of Equations

A system of equations is when you have multiple equations working together, and their solutions must satisfy all the equations simultaneously. In our case, the two equations form a system of nonlinear equations because they both include squared terms.

Here's a simple breakdown of how to solve a system of equations:

• Identify whether the system is linear or nonlinear. Here, they are nonlinear due to the squared terms.
• Use substitution or elimination methods to reduce the equations down to a simpler form.
• Work with one equation at a time to gradually find values for each variable that satisfy both equations.

The beauty of solving a system of equations is that it often reveals more insight into the relationship between the variables compared to solving each equation separately.

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest power of the variable is squared. In our exercise, one of the results of solving the system of equations is a quadratic equation of the form \(3y^2 - 3 = 0\).

To solve a quadratic equation, follow these steps:

• Try to simplify the equation, if possible, by dividing by any common factors. For example, dividing the quadratic by 3 gives us \(y^2 - 1 = 0\).
• Factor the equation, if applicable. Here, \(y^2 - 1 = 0\) can be factored as \((y - 1)(y + 1) = 0\).
• The solutions are the values that make each factor zero. So \(y = 1\) and \(y = -1\) are the solutions.

Quadratic equations can be solved using various methods like factoring, completing the square, or using the quadratic formula. Always double-check your solutions to ensure they satisfy the original problem conditions.