Problem 45 - College Algebra

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 of equations are \(x = 2, y = 1\) and \(x = -2, y = -1\)

Step 1: Translate Problem to Equations

The first step is to convert the given conditions into equations. \(x^2 - y^2 = 3\) is the translation of 'the difference between the squares of two numbers is 3'. The second condition can be translated as \(2x^2 + y^2 = 9\)

Step 2: Solving System of Equations

In order to solve for \(x\) and \(y\), the the system of equations can be solved. Start with the first equation: \(x^2 = y^2 + 3\). Substitute this into the second equation instead of \(x^2\): \(2(y^2 + 3) + y^2 = 9\). Simplify to find the value of \(y\)

Step 3: Find value of y

The equation becomes \(2y^2 + 6 + y^2 = 9\), simplifying further gives \(3y^2 + 6 = 9\), which reduces to \(3y^2 = 3\) and finally, \(y^2 = 1\). Thus, \(y = \pm 1\)

Step 4: Find value of x

Substitute \(y\) values into the first equation: \(x^2 = (\pm 1)^2 + 3\), simplifying gives \(x^2 = 1 + 3\), which gives \(x^2 = 4\). This means that \(x = \pm 2\)

Key Concepts

solving nonlinear systems

Solving nonlinear systems involves finding values for variables that satisfy multiple equations simultaneously. These systems contain equations that aren't linear and may include variables raised to powers higher than one or involve other nonlinear operators. In the problem at hand, we're dealing with a system consisting of two nonlinear equations:

• The difference between the squares of two numbers: \(x^2 - y^2 = 3\)
• Twice the square of the first number plus the square of the second: \(2x^2 + y^2 = 9\)

The challenge is to find values for \(x\) and \(y\) that make both equations true. This involves strategic manipulation and solving, typically requiring either algebraic or graphical techniques. Solving nonlinear systems can be more complex than solving linear systems, as they can have multiple solutions or no solutions at all. In this problem, we turn to substitution as a viable strategy for simplification and solution finding.

substitution method

The substitution method is a technique used to solve systems of equations by replacing one variable with an equivalent expression from another equation. This can simplify the system, making it easier to solve. Here's how it applies to our problem:
In the original system:

• Start with \(x^2 = y^2 + 3\) from the first equation \(x^2 - y^2 = 3\)
• Substitute this into the second equation \(2x^2 + y^2 = 9\). It becomes \(2(y^2 + 3) + y^2 = 9\)

This substitution effectively reduces the system to a single equation with one variable. Solving the resulting equation simplifies the process, allowing us to isolate the variable \(y\) first, and then solve for \(x\) using the values obtained. Substitution is a powerful tool particularly when the equations within a system share common variables that lead to an efficient simplification.

quadratic equations

Quadratic equations are polynomial equations of degree two, having the form \(ax^2 + bx + c = 0\). In this particular problem, both equations involve quadratics because the variables are squared.
For instance:

• The first equation involves a difference of squares: \(x^2 - y^2 = 3\)
• The substitution results in another quadratic that simplifies to find \(y\): \(3y^2 = 3\)

Solving these quadratics involves determining values of \(x\) and \(y\) that satisfy all conditions. Specifically, we set \(3y^2 = 3\) which yields \(y^2 = 1\), leading to potential solutions for \(y\) of \(\pm 1\). For the value of \(x\), we derive from \(x^2 - y^2 = 3\) which then becomes straightforward to handle once \(y\) is known.

difference of squares

The concept of the difference of squares is a specific algebraic identity applicable when you have two squared values subtracted from each other in the format \(a^2 - b^2 = (a - b)(a + b)\). Recognizing this pattern can significantly simplify calculations. In the exercise:

• The first equation \(x^2 - y^2 = 3\) directly aligns with the difference of squares identity.

The factored form is then \((x - y)(x + y) = 3\). Knowing this, one can consider possible simplifications or factor by this identity. Although actual factoring into linear terms isn’t needed in the solution method chosen here, awareness of this identity supports flexibility and potential alternative ways to dissect such problems. It also provides a broadened understanding of how equations might be manipulated and solved.