Problem 31 - Algebra and Trigonometry

Question

\(19-32\) . Find all solutions of the system of equations. $$ \left\\{\begin{array}{c}{\frac{2}{x}-\frac{3}{y}=1} \\\ {-\frac{4}{x}+\frac{7}{y}=1}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{5} \) and \( y = \frac{1}{3} \).

Step 1: Express the system of equations

We are given two equations: \(\frac{2}{x} - \frac{3}{y} = 1\) and \(-\frac{4}{x} + \frac{7}{y} = 1\). Our goal is to find values of \(x\) and \(y\) that satisfy both equations.

Step 2: Rearrangements

Let's rewrite both equations in terms of \( \frac{1}{x} \) and \( \frac{1}{y} \) to make substitution easier:\[2a - 3b = 1ewline-4a + 7b = 1ewline\]where \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \).

Step 3: Solving the system using elimination or substitution

To eliminate one of the unknowns, let's multiply the first equation by 2:\[4a - 6b = 2ewline-4a + 7b = 1ewline\]Now, add the two equations:\[(4a - 6b) + (-4a + 7b) = 2 + 1ewlineb = 3\]

Step 4: Solve for the second variable

Substitute \( b = 3 \) back into the first simplified equation:\[2a - 3(3) = 1 ewline2a - 9 = 1 ewline2a = 10 ewlinea = 5\]

Step 5: Translating back to original variables

Since \( a = \frac{1}{x} = 5 \) and \( b = \frac{1}{y} = 3 \), solving for \( x \) and \( y \) gives:\[x = \frac{1}{5}ewliney = \frac{1}{3}\]

Key Concepts

Substitution Method

The substitution method is a powerful tool for solving systems of equations. It involves solving one equation for one variable and then substituting that solution into another equation. This method is especially useful when one of the equations is easily solved for a single variable.

For example, in the given exercise, after expressing the equations in terms of new variables, namely \(a = \frac{1}{x}\) and \(b = \frac{1}{y}\), the substitution became straightforward. Once we found \(b = 3\) from one equation, this value was substituted back into the first equation to find \(a = 5\).

Here’s a quick reminder on the steps for substitution method:

• Solve one equation for one of its variables.
• Substitute this expression in the other equation, replacing the solved variable.
• Solve the resulting equation.
• Substitute back as needed to find the other variable.

It's a systematic approach and can simplify complex-looking problems.

Elimination Method

The elimination method is another effective technique for solving systems of equations. This involves adding or subtracting equations to eliminate one variable, allowing you to focus on the remaining one.

In the given exercise, after rewriting the equations in terms of \(a\) and \(b\), the elimination method was used by multiplying one of the equations to allow for easy addition or subtraction. By multiplying the first equation by 2 and adding it to the second, the \(a\) terms cancelled out, directly providing \(b\).

Here’s how you can use the elimination method:

• Adjust equations if necessary by multiplying both sides to align coefficients of one variable.
• Add or subtract the equations to eliminate one variable.
• Solve the resulting equation for the remaining variable.
• Back substitute to find other variables as needed.

The elimination method is perfect when the equations are set up nicely for addition or subtraction, or can easily be manipulated to align perfectly.

Rational Equations

Rational equations contain fractions with variables in the denominators. Solving these requires a careful approach, as the variables affect the entire fraction.

The given exercise initially presented the system of equations with rational expressions, like \(\frac{2}{x} - \frac{3}{y} = 1\). To make working with them simpler, a common approach is to express such equations in terms of their reciprocals \(\frac{1}{x}\) and \(\frac{1}{y}\).

Here's a straightforward path to tackling rational equations:

• Identify and rewrite each term with a variable,” finding a comparable and easier form like introducing reciprocals.
• Simplify the equations to make direct operations more feasible.
• Eliminate fractions by combining terms, substitution, or elimination.
• After solving, ensure the solutions are valid by substituting them back to the original equations.

Rational equations can seem tricky at times, but understanding the reciprocal relationships simplifies them significantly.