Problem 19 - Precalculus: Functions and Graphs

Question

Solve the system. $$\left\\{\begin{array}{l} \frac{2}{x}+\frac{3}{y}=-2 \\ \frac{4}{x}-\frac{5}{y}=1 \end{array} \quad\left(\text { Hint. Let } u=\frac{1}{x} \text { and } v=\frac{1}{y}\right)\right.$$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{22}{7} \) and \( y = -\frac{11}{5} \).

Step 1: Substitute new variables

The problem suggests using substitutions where \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Replace \( \frac{2}{x} \) with \( 2u \) and \( \frac{3}{y} \) with \( 3v \) in the first equation. Similarly, replace \( \frac{4}{x} \) with \( 4u \) and \( \frac{5}{y} \) with \( 5v \) in the second equation. The system becomes: \[\begin{cases}2u + 3v = -2 \4u - 5v = 1\end{cases}\]

Step 2: Solve system using elimination method

To eliminate one variable, multiply the first equation by 2, aiming to line up the coefficients of \( u \). This gives:\[ 4u + 6v = -4 \] Now subtract the second equation: \[ \begin{align*}(4u + 6v) - (4u - 5v) &= -4 - 1 \6v + 5v &= -5 \11v &= -5\end{align*}\] Solve for \( v \): \[ v = -\frac{5}{11} \]

Step 3: Solve for variable u

Substitute \( v = -\frac{5}{11} \) back into the first equation to find \( u \):\[ 2u + 3(-\frac{5}{11}) = -2 \] Simplify:\[ 2u - \frac{15}{11} = -2 \] Multiply through by 11 to clear the fraction:\[ 22u - 15 = -22 \] Add 15 to both sides:\[ 22u = -7 \] Divide by 22:\[ u = -\frac{7}{22} \]

Step 4: Substitute back to original variables

Remembering our substitutions, \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \), plug back the solved values: \[ \frac{1}{x} = -\frac{7}{22} \quad \text{and} \quad \frac{1}{y} = -\frac{5}{11} \] Thus, solving for \( x \) and \( y \): \[ x = -\frac{22}{7} \quad \text{and} \quad y = -\frac{11}{5} \]

Key Concepts

Elimination Method

The elimination method is a powerful technique used in solving systems of linear equations. This approach involves adding or subtracting equations to eliminate one of the variables, making it easier to solve for the remaining variable.
To effectively use this method, the key is to line up the coefficients of one of the variables so that when you add or subtract the equations, that variable is removed. This focuses your calculations on solving for one variable at a time, simplifying the entire process.

• Start by manipulating the equations to align the coefficients.
• Multiply equations by suitable numbers if necessary.
• Add or subtract the equations to eliminate one variable.
• Solve the resulting single-variable equation.
• Use the value found to solve for the other variable in one of the original equations.

In the given exercise, we use the elimination method to solve for one variable by first aligning the coefficients for the variable \( u \). By multiplying the first equation by 2, we make the coefficients of \( u \) in both equations equal (4), allowing us to subtract the second from the first, thereby eliminating \( u \) and solving directly for \( v \). With \( v \) found, it's straightforward to solve for \( u \) using substitution back into one of the original equations.

Substitution Method

The substitution method is a straightforward procedure used to solve systems of equations by replacing one variable with an expression involving the other variable. This method is highly valuable when the equations can be easily manipulated to solve for one variable in terms of the others.
Here's a simple way to apply the substitution method:

• Start by solving one of the equations for one variable in terms of the other. This step often requires isolating the variable on one side of the equation.
• Substitute the expression obtained in the previous step into the other equation. This effectively reduces the system to a single equation with one variable.
• Solve the resulting equation for the variable.
• Back-substitute the solution into the expression obtained in the first step to find the second variable.

In our exercise, substitution is cleverly introduced by redefining the original equations with new variables \( u \) and \( v \). This remarkably simplifies the system, allowing us to approach the solution systematically. Once we determine the values of \( u \) and \( v \), we return to the original definitions of these variables, \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \), to solve for \( x \) and \( y \), completing the solution.

Rational Equations

Rational equations are equations that involve fractions whose numerators and denominators are polynomials. These types of equations can pose unique challenges, primarily due to the presence of variables in the denominators. When dealing with rational equations, the fundamental goal is typically to eliminate the fractions to simplify the solving process.
You might follow these basic steps to tackle rational equations:

• Identify the least common denominator (LCD) of all the fractions involved.
• Multiply every term by the LCD to clear the fractions.
• With the fractions eliminated, what remains is often a polynomial equation that can be handled using more straightforward algebraic techniques.
• Solve the resulting equation, being mindful to check for extraneous solutions since the original equation was rational.

In the provided problem, the rational equations involve fractions like \( \frac{2}{x} \) and \( \frac{3}{y} \). By making smart substitutions, \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \), we transform these rational aspects into linear equations, which are easier to manage mathematically. This approach not only simplifies the task but also opens the path to applying linear solving techniques such as substitution and elimination effectively.