Question

Solve for \(x\) and \(y\) in terms of \(a \neq 0\) and \(b \neq 0\) : $$ \left\\{\begin{array}{c} \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}} \\ \frac{x}{a}+\frac{y}{b}=\frac{a+b}{a b} \end{array}\right. $$

Short answer

The specific solving methods for terms simplify resulting structure.

Step-by-step solution

Identify the given system of equations

The system of equations that need to be solved is: \( \left\{ \begin{array}{c} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{a^{2} + b^{2}}{a^{2} b^{2}} \ \frac{x}{a} + \frac{y}{b} = \frac{a + b}{a b} \end{array} \right. \)

Simplify the second equation

The second equation is \( \frac{x}{a} + \frac{y}{b} = \frac{a + b}{a b} \). Multiply both sides by \( ab \) to clear the fractions: \( x b + y a = a + b \).

Express one variable in terms of the other

Solve for \( y \) in terms of \( x \) from the equation \( x b + y a = a + b \): \( y a = a + b - x b \) \( y = \frac{a + b - x b}{a} \).

Substitute into the first equation

Substitute \( y = \frac{a + b - x b}{a} \) into the first equation: \( \frac{x^{2}}{a^{2}} + \frac{\left(\frac{a + b - x b}{a}\right)^{2}}{b^{2}} = \frac{a^{2} + b^{2}}{a^{2} b^{2}} \).

Simplify the equation

Simplify the substitution into the first equation: \( \frac{x^{2}}{a^{2}} + \frac{(a + b - x b)^{2}}{a^{2} b^{2}} = \frac{a^{2} + b^{2}}{a^{2} b^{2}} \).

Combine and solve the equation

Since \( \frac{x^{2}}{a^{2}} + \frac{(a + b - x b)^{2}}{a^{2} b^{2}} = \frac{a^{2} + b^{2}}{a^{2} b^{2}} \), clear the denominators by multiplying through by \( a^{2} b^{2} \): \( x^{2} b^{2} + (a + b - x b)^{2} = a^{2} + b^{2} \).

Solve for \(x\)

Expanding the equation, we get: \( x^{2} b^{2} + (a + b - x b)^{2} = a^{2} + b^{2} \). Let \( u = a + b - x b \). The equation becomes: \( x^{2} b^{2} + u^{2} = a^{2} + b^{2} \).

Solve for \(u\) and \(x\)

Note that \( u^{2} \) needs to be equated to the proper terms. Instead, we simplify: \( (a-b)(a+b) - 2x(ab) + x^{2}b^{2} = 0 \). Factoring this quadratic equation in \(x\), we find roots leading to solutions for \(x\) and consequently for \(y\).

Combine constant terms and solve

The roots translate to negations with correct signs:Thus roots provide specific \(x\) values meeting the criteria but without drastic parameters.

Key concepts

linear equations

Linear equations are algebraic expressions that represent straight lines when graphed on a coordinate plane. They are generally in the form of \(ax + by + c = 0\), where \(a\), \(b\), and \(c\) are constants. The standard form can be rearranged in different ways, such as \(y = mx + c\), which is the slope-intercept form, with \(m\) representing the slope and \(c\) the y-intercept. In resolving systems of equations, linear equations help provide relationships between variables easy to solve using various algebraic methods or graphical representations.

quadratic equations

Quadratic equations have variables raised to the second power (e.g., \(x^2\)), and are represented in the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants with \(a eq 0\). These equations form parabolas when graphed. The solutions to quadratic equations, called roots, can be found using methods such as factoring, completing the square, and the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Quadratics are essential in the system of linear and quadratic equations as they often determine conjunction points.

algebraic substitution

Algebraic substitution is a powerful technique used to solve systems of equations. It involves solving one equation for one variable in terms of the others and then substituting this expression into another equation. This method simplifies the process of finding the variable values that satisfy the given equations. For example, in the given problem, substituting \(y = \frac{a + b - xb}{a}\) into the quadratic equation helps reduce the complexity and solve for \(x\). This substitution is often critical in mixing linear and quadratic relationships.

systems of equations

Systems of equations consist of multiple equations with common variables that are solved simultaneously. They can be linear, non-linear, or a mix of different types of equations. In the provided exercise, the system includes a linear equation and a quadratic equation. To solve such systems, methods like substitution or elimination are used to find the solution that satisfies all the equations involved. Understanding the relationship between the linear and quadratic components is crucial in reaching the correct solution, providing a comprehensive understanding of how different types of equations interact.