Problem 64 - 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 \(b=\frac{1}{y}\) $$ \left\\{\begin{array}{l} \frac{3}{x}-\frac{2}{y}=-30 \\ \frac{2}{x}-\frac{3}{y}=-30 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{1}{6} \) and \( y = \frac{1}{6} \).

Step 1: Introduce the Substitutions

Let \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \). Substitute these into the system of equations: \( 3a - 2b = -30 \) and \( 2a - 3b = -30 \).

Step 2: Solve the Linear System for a and b

Arrange the equations: \( \begin{align*} 3a - 2b &= -30 \ 2a - 3b &= -30 \end{align*} \). Use elimination or substitution to solve for \( a \) and \( b \). Multiply the first equation by 2 and the second by 3 to eliminate \( a \): \( \begin{align*} 6a - 4b &= -60 \ 6a - 9b &= -90 \end{align*} \). Subtract the second equation from the first to get: \( 5b = 30 \). Solve for \( b \): \( b = 6 \). Substitute \( b = 6 \) into the first original equation: \( 3a - 2(6) = -30 \), which simplifies to \( 3a - 12 = -30 \) and solve for \( a \): \( 3a = -18 \) giving \( a = -6 \).

Step 3: Determine x and y

Recall the substitutions: \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \). Solve for \( x \) and \( y \): since \( a = -6 \), it implies \( x = \frac{1}{-6} = -\frac{1}{6} \). Since \( b = 6 \), it implies \( y = \frac{1}{6} \).

Step 4: Verify the Solution

Substitute \( x = -\frac{1}{6} \) and \( y = \frac{1}{6} \) back into the original equations to verify. For the first equation: \( \frac{3}{x} - \frac{2}{y} = 3\times (-6) - 2\times 6 = -18 - 12 = -30 \) which is correct. For the second equation: \( \frac{2}{x} - \frac{3}{y} = 2\times (-6) - 3\times 6 = -12 - 18 = -30 \) which is also correct.

Key Concepts

Substitution Method

The substitution method is a powerful tool used to solve systems of equations. In this method, you want to express one variable in terms of another in a simple form to substitute it back into the equation, simplifying it.
In the given exercise, we substitute by letting \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \). This transforms our original system of equations into more straightforward linear equations, which can be easier to handle:

• \( 3a - 2b = -30 \)
• \( 2a - 3b = -30 \)

By making these substitutions, the complex fractions involving \( x \) and \( y \) are avoided early in the problem-solving process. This method shows that substitutions can significantly simplify problems that initially seem cumbersome.

Algebraic Manipulation

After substituting variables, algebraic manipulation becomes crucial to solving the equations. Here, manipulation involves using operations like addition, subtraction, multiplication, or division to eliminate one variable and solve for the other.
The core of algebraic manipulation lies in maintaining equality while simplifying equations. Our problem shows how to use these techniques effectively:

• Start by aligning the equations for elimination: \( 3a - 2b = -30 \) and \( 2a - 3b = -30 \).
• Multiply equations to align terms creatively (e.g., both equations multiplied to cancel out \( a \)).
• Effectively subtract one from the other, simplifying to find \( b \).

After isolating a variable, the rest is about reverse operations to achieve both solutions \( a \) and \( b \), then revert back to \( x \) and \( y \).

Linear Equations Solution Methods

Solving systems of linear equations usually involves methods like substitution or elimination. Each method has its uses, depending on the problem.
The method of elimination is particularly useful when the equations are fairly straightforward, and you can easily cancel out a variable by adding or subtracting equations. Substitution is used here to transform variables, making complex equations more approachable.

• Substitution can simplify equations by reducing them to involve a single variable at a time.
• Elimination ultimately helps resolve coefficient mismatches for easy calculation.

These methods, when combined, offer a systematic approach to solving a system of equations step-by-step, ensuring you can find the values of the variables without guessing.