Problem 54 - Elementary Algebra

Question

. How can you tell by inspection that \(\frac{x}{x-4}=\frac{4}{x-4}\) has no solution?

Step-by-Step Solution

Answer

The equation has no solution because substituting \( x = 4 \) leads to division by zero.

Step 1: Identify the Equation

The given equation is \( \frac{x}{x-4} = \frac{4}{x-4} \). Notice that both sides of the equation have the same denominator \( x - 4 \).

Step 2: Cross-Multiply

To eliminate the fractions, we can cross-multiply the equation: \( x \times (x - 4) = 4 \times (x - 4) \).

Step 3: Simplify Both Sides

Distribute the terms on both sides: \( x(x - 4) = 4(x - 4) \), which simplifies to \( x^2 - 4x = 4x - 16 \).

Step 4: Set the Equation to Zero

Bring all terms to one side of the equation: \( x^2 - 4x - 4x + 16 = 0 \) becomes \( x^2 - 8x + 16 = 0 \).

Step 5: Factor the Quadratic

Recognize that this equation \( x^2 - 8x + 16 = 0 \) can be factored to \( (x - 4)^2 = 0 \).

Step 6: Determine the Solution

Solving \( (x - 4)^2 = 0 \) gives \( x = 4 \) as the only possible solution.

Step 7: Check the Solution in the Original Equation

Substitute \( x = 4 \) into the original equation: \( \frac{4}{4 - 4} = \frac{4}{4 - 4} \), which leads to division by zero, indicating that \( x = 4 \) is not a valid solution.

Step 8: Conclusion

Since both sides of the equation result in division by zero when \( x = 4 \), there is no solution to the equation.

Key Concepts

Cross Multiplication

Cross multiplication is a powerful technique used to solve equations where two fractions are set equal to each other. It allows you to transform the equation by eliminating the fractions which makes it easier to solve. For example, if you have an equation like \( \frac{a}{b} = \frac{c}{d} \), you can solve it by cross-multiplying, which means multiplying the numerator of one fraction by the denominator of the other and vice versa. In this case, you'd multiply as follows:

• \( a \times d = b \times c \)

When using cross-multiplication, it's key to remember that you are essentially using the property of equality, stating that if two fractions are equal, their cross-products must also be equal. This method can simplify the process significantly by removing the need to deal with fractions directly.
In our specific exercise, we applied cross-multiplication to the equation \( \frac{x}{x-4} = \frac{4}{x-4} \), resulting in \( x(x - 4) = 4(x - 4) \). This let us work on simplifying and factoring without fractions.

Division by Zero

The issue of division by zero is a fundamental concept in mathematics that can lead to undefined expressions. Division by zero occurs when a number is divided by another number that is zero. Such an operation is not defined in traditional arithmetic because it's impossible to calculate a definite number when the divisor is zero.
If you think about division by zero, the main problem arises from trying to distribute something (say, a number 'a') into zero equal parts, which simply does not make sense.
In the exercise \( \frac{x}{x-4} = \frac{4}{x-4} \), replacing \( x \) with the denominator value leads to division by zero. Initially, cross-multiplying and simplifying the fractions led us to the potential solution \( x = 4 \). However, plugging \( x = 4 \) back into the original equation creates a scenario of division by zero, as the denominator \( x - 4 \) becomes zero. This confirms that there is no valid solution to the equation.

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). These equations can be solved by factoring, using the quadratic formula, or completing the square. The key is recognizing the characteristic 'squared' term \( x^2 \). For the equation to be quadratic, the highest exponent of the variable is 2.

• Factoring involves expressing the quadratic equation as a product of two binomials.
• The quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) provides a direct way to find solutions.
• Completing the square offers another technique, transforming the equation to \( (x - p)^2 = q \).

In our exercise, the equation \( x^2 - 8x + 16 = 0 \) was easily factorable to form \( (x - 4)^2 = 0 \), suggesting a solution of \( x = 4 \). One might hastily conclude a solution exists by factoring but always needs to check for division by zero from the original formulation. This reaffirms the importance of considering all properties of the equation before concluding the validity of solutions.