Problem 62 - Intermediate Algebra

Question

Solve each system. To do so, substitute a for \(\frac{1}{x}\) and \(b\) for \(\frac{1}{y}\) and solve for a and \(b\). Then find \(x\) and \(y\) using the fact that \(a=\frac{1}{x}\) and \(b=\frac{1}{y}\) $$ \left\\{\begin{array}{l} \frac{1}{x}+\frac{1}{y}=\frac{9}{20} \\ \frac{1}{x}-\frac{1}{y}=\frac{1}{20} \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = 2\) and \(y = -20\).

Step 1: Substitute Variables

Substitute \(a = \frac{1}{x}\) and \(b = \frac{1}{y}\) into the system of equations. This transforms the system into:1. \( a + b = \frac{9}{20} \)2. \( a - b = \frac{1}{20} \)

Step 2: Solve for a and b

To eliminate \(b\), add the two equations:\( (a + b) + (a - b) = \frac{9}{20} + \frac{1}{20} \)This simplifies to \(2a = \frac{10}{20}\), so \(a = \frac{1}{2}\).Now substitute \(a = \frac{1}{2}\) back into the first equation:\(\frac{1}{2} + b = \frac{9}{20}\)Subtract \(\frac{1}{2}\) from both sides: \(b = \frac{9}{20} - \frac{10}{20} = -\frac{1}{20}\).

Step 3: Find x and y

Now that we have \(a = \frac{1}{2}\) and \(b = -\frac{1}{20}\), solve for \(x\) and \(y\):- Since \(a = \frac{1}{x}\), then \(x = \frac{1}{a} = 2\).- Since \(b = \frac{1}{y}\), then \(y = \frac{1}{b} = -20\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to simplify and solve systems of equations. Here's how it works:

1. Choose one equation and solve for one variable in terms of another.
2. Substitute this expression into the other equations.
3. Simplify and solve the resulting equation to find the value of one variable.
4. Substitute back to find the remaining variables.

This method is particularly useful when dealing with linear equations because it reduces the system into a single equation in one variable. In our exercise, we replace \( \frac{1}{x} \) with \( a \) and \( \frac{1}{y} \) with \( b \). This transforms the system into a simpler form,
- \( a + b = \frac{9}{20} \)
- \( a - b = \frac{1}{20} \)
allowing us to solve for \( a \) and \( b \) more easily.

Through substitution, complex equations become manageable, striking the challenge down into individual, easily solvable algebraic problems.

Solving Linear Equations

Solving linear equations is foundational in math and crucial in handling systems like these. A linear equation represents a straight line when plotted on a graph and has no variables raised to powers higher than one.

The goal in solving them is to isolate the variable on one side of the equation.
Let's explore the steps applied in the exercise:

- After substituting \( a \) and \( b \) into the system, we had two linear equations:
- \( a + b = \frac{9}{20} \)
- \( a - b = \frac{1}{20} \)
- We can then linearly combine or manipulate these equations by adding them together.
- The addition removes \( b \) and doubles \( a \). Solving \( 2a = \frac{10}{20} \) gives us \( a = \frac{1}{2} \).

This technique of systematically isolating variables is central to solving linear equations efficiently, allowing for straightforward solutions.

Inverse Operations

Inverse operations are like the undo buttons of math. They reverse actions to help solve equations.

In our exercise, inverse operations play a key role. Once we identify \( a \) and \( b \), we need to find \( x \) and \( y \).

Here’s where inverse operations shine:

• Since \( a = \frac{1}{x} \), we use the inverse operation (reciprocal) to find \( x = \frac{1}{a} = 2 \).
• Similarly, for \( b = \frac{1}{y} \), the reciprocal gives us \( y = \frac{1}{b} = -20 \).

Inverses simplify finding original variables, converting complex fractions into simple arithmetic. Remember, when dealing with any function, its inverse will always bring you back to the starting point.