Chapter 11: Problem 29
Equations with Unknown in Denominator. \(\frac{x-3}{x+2}=\frac{x+4}{x-5}\)
Short Answer
Expert verified
The value of x is \( \frac{1}{2} \) or \(0.5\
Step by step solution
01
Cross-Multiply
Since \(\frac{a}{b} = \frac{c}{d}\) implies \(a \cdot d = b \cdot c\) when \(b \eq 0\) and \(d \eq 0\), cross-multiply to remove the fractions: \( (x - 3)(x - 5) = (x + 4)(x + 2) \).
02
Distribute and Expand both sides
Distribute the products on both sides of the equation: \( x^2 - 5x - 3x + 15 = x^2 + 2x + 4x + 8 \).
03
Combine Like Terms
Combine like terms on each side of the equation: \( x^2 - 8x + 15 = x^2 + 6x + 8 \).
04
Move all terms to one side
Subtract \(x^2\), \(6x\), and \(8\) from both sides to get all terms on one side and the equation equal to zero: \( -8x - 6x + 15 - 8 = 0 \).
05
Combine Like Terms again
Combine the remaining like terms to simplify the equation: \( -14x + 7 = 0 \).
06
Solve for x
Add \(14x\) to both sides to isolate x: \( x = \frac{7}{14} \) and then simplify the fraction: \( x = \frac{1}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross-Multiplication
Cross-multiplication is a technique used to solve equations that have fractions with variables in the denominator. It makes solving these rational equations more straightforward by eliminating the fractions altogether. To cross-multiply, you take the numerator of the first fraction and multiply it by the denominator of the second fraction and set it equal to the product of the numerator of the second fraction and the denominator of the first fraction.
In our example, we cross-multiply \(\frac{x-3}{x+2}=\frac{x+4}{x-5}\) to get \( (x - 3)(x - 5) = (x + 4)(x + 2) \). It is essential to remember that cross-multiplication is only valid if neither denominator is zero since division by zero is undefined.
In our example, we cross-multiply \(\frac{x-3}{x+2}=\frac{x+4}{x-5}\) to get \( (x - 3)(x - 5) = (x + 4)(x + 2) \). It is essential to remember that cross-multiplication is only valid if neither denominator is zero since division by zero is undefined.
Distributive Property
The distributive property is a fundamental algebraic property used in simplifying expressions and equations involving multiplication over addition or subtraction. It states that \(a(b + c) = ab + ac\). This property is applied to expand expressions by eliminating parentheses.
During the second step of the solution, we use the distributive property to expand both sides of the cross-multiplied equation: \(x^2 - 8x + 15 = x^2 + 6x + 8\). Careful distribution ensures that each term is multiplied correctly, allowing us to proceed to the next steps in simplifying the equation.
During the second step of the solution, we use the distributive property to expand both sides of the cross-multiplied equation: \(x^2 - 8x + 15 = x^2 + 6x + 8\). Careful distribution ensures that each term is multiplied correctly, allowing us to proceed to the next steps in simplifying the equation.
Combining Like Terms
Combining like terms is a process that involves consolidating terms with the same variable raised to the same power into a single term. This simplification step is crucial in solving algebraic equations efficiently. In the example, this process is used twice.
First, we combine like terms after distributing: \(x^2 - 5x - 3x + 15\) simplifies to \(x^2 - 8x + 15\), and \(x^2 + 2x + 4x + 8\) simplifies to \(x^2 + 6x + 8\). Later, after moving all terms to one side, we combine \( -8x - 6x\) to get \( -14x\), and \(15 - 8\) simplifies to 7, which brings us to the simplified form \( -14x + 7 = 0\). Combining like terms is a crucial step to arrive at a more manageable expression to solve for the variable.
First, we combine like terms after distributing: \(x^2 - 5x - 3x + 15\) simplifies to \(x^2 - 8x + 15\), and \(x^2 + 2x + 4x + 8\) simplifies to \(x^2 + 6x + 8\). Later, after moving all terms to one side, we combine \( -8x - 6x\) to get \( -14x\), and \(15 - 8\) simplifies to 7, which brings us to the simplified form \( -14x + 7 = 0\). Combining like terms is a crucial step to arrive at a more manageable expression to solve for the variable.
Equations with Variables in the Denominator
Solving equations with variables in the denominator requires special attention since you can't divide by zero. In such cases, the domain of the variable should exclude any values that would make the denominator zero.
In our exercise, the presence of variables in the denominators of both fractions prompts us to approach solving the equation cautiously. Before implementing cross-multiplication, we acknowledge that \(x+2\) and \(x-5\) cannot be zero. Therefore, x cannot be -2 or 5 in the solution. After cross-multiplication and following the subsequent steps, we find that \(x = \frac{1}{2}\), which is within the allowable domain. Always examine the solution against the original equation's domain to ensure it is valid.
In our exercise, the presence of variables in the denominators of both fractions prompts us to approach solving the equation cautiously. Before implementing cross-multiplication, we acknowledge that \(x+2\) and \(x-5\) cannot be zero. Therefore, x cannot be -2 or 5 in the solution. After cross-multiplication and following the subsequent steps, we find that \(x = \frac{1}{2}\), which is within the allowable domain. Always examine the solution against the original equation's domain to ensure it is valid.
