Problem 10 - Algebra 1

Question

Solve the equation. Remember to check for extraneous solutions. $$\frac{4}{x(x+1)}=\frac{3}{x}$$

Step-by-Step Solution

Answer

The solution to the given equation is \(x=\frac{1}{3}\).

Step 1: Cross Multiplication

In order to avoid the fractions, apply cross multiplication. So you will multiply each side of the equation by \(x\) and by \(x(x+1)\). Resulting in \(4x=3x(x+1)\)

Step 2: Simplify and Rearrange the Equation

Now, simplify the right side of equation by distributing 3x, you get \(4x=3x^2+3x\). To simplify, subtract 4x from each side to set the equation equal to zero resulting in \(3x^2+3x-4x=0\) which simplifies to \(3x^2-x=0\).

Step 3: Factor the Equation

To solve the quadratic equation, we need to factor it first. This simplifies to \(x(3x-1)=0\).

Step 4: Solve for x

The value of x is then obtained by setting each factor equal to zero and solving: \(x=0\) and \((3x-1)=0\) resulting in \(x=0\) and \(x=\frac{1}{3}\).

Step 5: Check for Extraneous Solutions

Substitute each solution back into the original equation: with \(x=0\), the original equation becomes undefined because it will result in dividing by zero. So, \(x=0\) is an extraneous solution. With \(x=\frac{1}{3}\), the left side of the original equation equals the right, meaning it is a valid solution.

Key Concepts

Cross Multiplication

Cross multiplication is a powerful technique used to solve equations involving fractions. It involves multiplying the numerator of each fraction by the denominator of the other fraction.

• In our exercise, we have two fractions: \( \frac{4}{x(x+1)} \) and \( \frac{3}{x} \). By cross-multiplying, we eliminate the fractions.
• This means we multiply the left side of the equation by the denominator of the right side, which is \( x \), and the right side by the denominator of the left side, which is \( x(x+1) \).
• The equation then becomes \( 4x = 3x(x+1) \), with no fractions involved.

Cross multiplication helps in simplifying complex rational equations, making it a fundamental step in solving them.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two, typically in the form \( ax^2 + bx + c = 0 \).

• After we apply cross multiplication in our exercise, the equation simplifies further: \( 4x = 3x^2 + 3x \).
• Rearranging this equation into standard quadratic form, we subtract \(4x\) from both sides, resulting in \( 3x^2 + 3x - 4x = 0 \).
• Simplifying further, we get \( 3x^2 - x = 0 \).

Quadratic equations can often be solved by factoring, completing the square, or using the quadratic formula.

Factoring

Factoring involves breaking down an equation into simpler expressions that, when multiplied together, give the original equation.

• For the quadratic equation \(3x^2 - x = 0\), factoring is performed by finding common factors in each term.
• The factorization of \(3x^2 - x = 0\) is \(x(3x - 1) = 0\).

When you have a factored equation like this, you can find the solutions by setting each factor equal to zero. This is the basis for solving quadratic equations by factoring.

Extraneous Solutions

Extraneous solutions are solutions that emerge in the process of solving an equation but do not satisfy the original equation. This often occurs when solving rational equations.

• In our exercise, the solutions derived from the factored equation are \(x = 0\) and \(x = \frac{1}{3}\).
• Checking \(x = 0\) involves substituting it back into the original equation \( \frac{4}{x(x+1)} = \frac{3}{x} \). This led to division by zero, making it invalid.
• Checking \(x = \frac{1}{3}\) confirmed it as a valid solution, as it did not result in any mathematical errors.

Always verify each solution by substituting back into the original equation to identify any extraneous solutions, ensuring your final answers are correct.