Problem 61 - 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 \(b=\frac{1}{y}\) $$ \left\\{\begin{array}{l} \frac{1}{x}+\frac{1}{y}=\frac{5}{6} \\ \frac{1}{x}-\frac{1}{y}=\frac{1}{6} \end{array}\right. $$

Step-by-Step Solution

Answer

Solutions are \( x = 2 \) and \( y = 3 \).

Step 1: Substitute Variables

Let \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \). Substitute these into the system of equations to replace \( \frac{1}{x} \) and \( \frac{1}{y} \) with \( a \) and \( b \). The system becomes: \[ \begin{cases} a + b = \frac{5}{6} \ a - b = \frac{1}{6} \end{cases} \]

Step 2: Solve for 'a' and 'b'

Add the two equations to eliminate \( b \):\[ (a + b) + (a - b) = \frac{5}{6} + \frac{1}{6} \]This simplifies to:\[ 2a = 1 \]Dividing both sides by 2 gives:\[ a = \frac{1}{2} \]

Step 3: Solve for 'b'

Use the equation \( a + b = \frac{5}{6} \) and substitute \( a = \frac{1}{2} \):\[ \frac{1}{2} + b = \frac{5}{6} \]Subtract \( \frac{1}{2} \) from both sides:\[ b = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3} \]

Step 4: Find 'x' and 'y'

Recall that \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \):For \( a \):\[ \frac{1}{x} = \frac{1}{2} \Rightarrow x = 2 \]For \( b \):\[ \frac{1}{y} = \frac{1}{3} \Rightarrow y = 3 \]

Key Concepts

Substitution Method

The substitution method is a crucial tool for solving systems of equations, especially when we have reciprocal relationships involved, like in our current system. The core idea is simple: replace one variable with an equivalent expression involving another variable that you can solve for first.
Take a step-by-step approach:

• First, identify an equation that can be easily manipulated to isolate a variable.
• Substitute this expression into the other equation(s) within the system.
• Solve the resulting equation to find the value of a single variable.

This method is particularly helpful when variables involve fractions or reciprocals, as directly solving for the variable would be challenging. By utilizing substitution, you can simplify complex problems into more easily manageable steps.

Solving Linear Equations

When solving linear equations, especially in a system, clarity and structure are essential. Linear equations are equations of the first degree, meaning they include terms that are constants or multiplied by variables raised to the power of one. In our system:

• We use operations like addition or subtraction to eliminate variables strategically.
• One equation can be added to or subtracted from another to solve for one of the variables.
• It's important to perform the same operation on both sides of the equation to maintain equality.

In this particular problem, by adding the equations together, we eliminated one variable (in this case, 'b') to solve for 'a.' Once 'a' is found, substituting back lets you solve for the other variable 'b,' completing the system solution efficiently.

Reciprocal Relationships

Reciprocal relationships involve the concept of a reciprocal, where two numbers multiply to one (such as \( a \) is the reciprocal of \( x \) because \((a = \frac{1}{x})\). These are particularly intriguing in equations because they offer a way to deal with division via multiplication.

• The reciprocal of a fraction \((\frac{1}{x})\) simplifies complex division problems by turning into simple multiplications.
• Finding reciprocal relationships can help in rewriting equations to substitute variables easily.
• This concept not only simplifies equations for substitution but also smoothly leads to solving once variables are isolated.

In systems of equations, identifying reciprocal relationships allows you to convert intricate equations into a more familiar linear format, thus simplifying the entire solution process. When you ultimately solve for the equations' variables, converting back from the reciprocals to the original format helps find values for the system's initial expressions.