Problem 20 - Precalculus: Functions and Graphs

Question

Solve the system. $$\left\\{\begin{array}{l} \frac{3}{x-1}+\frac{4}{y+2}=2 \\ \frac{6}{x-1}-\frac{7}{y+2}=-3 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{47}{2}\) and \(y = \frac{1}{7}\).

Step 1: Recognize the Variables

Let's first identify the variables given in the system of equations, which are \(x\) and \(y\). These variables are found within the denominators of the fractions.

Step 2: Identifying the Common Denominator

The equations involve the terms \(\frac{3}{x-1}\), \(\frac{4}{y+2}\), \(\frac{6}{x-1}\), and \(\frac{7}{y+2}\). Notice that they share common expressions in the denominators: \(x-1\) and \(y+2\). This suggests the substitution approach using \(a = x-1\) and \(b = y+2\).

Step 3: Substitute Variables

Replace \(x-1\) with \(a\) and \(y+2\) with \(b\) in both equations. The system becomes: 1. \(\frac{3}{a} + \frac{4}{b} = 2\) 2. \(\frac{6}{a} - \frac{7}{b} = -3\).

Step 4: Solve the First Equation for One Variable

From the first equation \(\frac{3}{a} + \frac{4}{b} = 2\), solve for \(a\) in terms of \(b\): - Multiply through by \(ab\) to clear fractions: \(3b + 4a = 2ab\).- Rearrange: \(3b = 2ab - 4a\). - Factor \(a\): \(3b = a(2b - 4)\).- Isolate \(a\): \(a = \frac{3b}{2b - 4}\).

Step 5: Solve the Second Equation for the Other Variable

Using the second equation \(\frac{6}{a} - \frac{7}{b} = -3\), solve for \(b\):- Multiply through by \(ab\) to clear fractions: \(6b - 7a = -3ab\).- Rearrange: \(6b = -3ab + 7a\).- Factor \(a\): \(6b = a(-3b + 7)\).- Isolate \(a\): \(a = \frac{6b}{-3b + 7}\).

Step 6: Set Equations Equal to Each Other

Since \(a = \frac{3b}{2b - 4}\) and \(a = \frac{6b}{-3b + 7}\), equate them:- \(\frac{3b}{2b - 4} = \frac{6b}{-3b + 7}\).

Step 7: Cross-Multiply and Simplify

Cross-multiply to solve for \(b\):- \(3b(-3b + 7) = 6b(2b - 4)\)- Simplify both sides: \(-9b^2 + 21b = 12b^2 - 24b\).- Combine like terms: \(21b + 24b = 12b^2 + 9b^2 \).- Simplify further: \(45b = 21b^2\).- Rearrange: \(21b^2 - 45b = 0\).- Factor: \(b(21b - 45) = 0\).- Solve: \(b = 0\) or \(21b = 45\).

Step 8: Solve for b

Solve \(21b = 45\): - \(b = \frac{45}{21} = \frac{15}{7}\).- Ensure \(b eq 0\) since division by zero is undefined for original expressions.

Step 9: Substitute Back to Find x and y

Knowing \(b = \frac{15}{7}\), substitute back into \(y+2 = b\) to find \(y\):- \(y + 2 = \frac{15}{7}\) implies \(y = \frac{15}{7} - \frac{14}{7} = \frac{1}{7}\).Using \(a = \frac{3b}{2b - 4}\), substitute back to find \(x\):- \(a = x - 1\) implies \(x = a + 1\).- Since \(a = \frac{3(\frac{15}{7})}{2(\frac{15}{7}) - 4} = \frac{45}{2 \times 15 - 28} = \frac{45}{2}\).- \(x = \frac{45}{2} + 1 = \frac{45}{2} + \frac{2}{2} = \frac{47}{2}\).

Key Concepts

Substitution Method

The substitution method is a technique often used to solve systems of equations, specifically when you have common elements that allow simplification. In this exercise, we use the substitution method to handle the fractions. By identifying common denominators for both fractions, we can simplify our systems significantly. Let's substitute by introducing new variables. In these equations, let \( a = x-1 \) and \( b = y+2 \). These substitutions focus on the expressions found in the denominators. This step changes the original complex equations into simpler forms that are easier to manage:

• \( \frac{3}{a} + \frac{4}{b} = 2 \)
• \( \frac{6}{a} - \frac{7}{b} = -3 \)

By working with these simpler forms, it becomes easier to manipulate and solve for the new variables \( a \) and \( b \), which can then lead us back to the original variables \( x \) and \( y \). This method is particularly useful when dealing with complicated equations or when direct substitution for one variable is cumbersome.

Rational Equations

Rational equations involve fractions containing polynomials in either the numerator, the denominator, or both. When working with rational equations, like in our exercise, it's crucial to identify a common factor in the denominator when possible, and then attempt to clear the fractions. This clarity often leads to simpler algebraic expressions. However, watch out for potential pitfalls, such as division by zero, which means \( x = 1 \) or \( y = -2 \) would lead to undefined expressions in the original denominators:\( x-1 \) and \( y+2 \).

• Identify the potential restrictions caused by denominators
• Simplify the problem by eliminating fractions step by step

The key is to multiply through each equation by the least common denominator of all the rational expressions involved. This transforms complex equations into a more straightforward form that's often easier to manage.

Fractional Equations

Fractional equations are similar to rational equations but focus more on expressions within each part of a fraction instead of the entire equation. In these kinds of problems, it is crucial to maintain a keen eye on simplifying the terms on both sides of the equation. The goal here is to eliminate the fractions by multiplying through the entire equation by the least common denominator. In our system,

• Multiply to remove denominators: \( ab(\frac{3}{a} + \frac{4}{b} = 2) \)
• The result: \( 3b + 4a = 2ab \)

By eliminating the fractions, you can focus on pure algebraic manipulation without the complicating factors of division errors, such as the aforementioned indeterminate forms or undefined values.

Algebraic Manipulation

Algebraic manipulation includes using operations to rearrange and solve equations. This practice involves moving terms from one side of the equation to the other to isolate the variable of interest. In the given solution, we apply algebraic manipulation after clearing the fractions:

• Start with expressions like \( 3b = 2ab - 4a \)
• Rearrange terms: \( 3b = a(2b - 4) \)

By rearranging terms, we aim to express one variable in terms of others, simplifying our task of finding specific values. Useful techniques include factoring out common terms and simplifying fractions.Breaking down each step carefully makes the process manageable and provides clarity, allowing each move towards a solution to be checked for accuracy.