Problem 58 - Algebra 2

Question

Solve each equation. Check your solution. $$ \frac{x}{x+1}-\frac{x}{x-3}=9 $$

Step-by-Step Solution

Answer

The solution to the equation \( \frac{x}{x+1} - \frac{x}{x-3} = 9 \) is \( x_1 \) and \( x_2 \), where \( x_1 \) and \( x_2 \) are the values found from the Quadratic Formula.

Step 1: Clear the denominators

To clear the denominators, we'll find a common denominator. Here, it is \( (x+1)(x-3) \). Multiply each term of the equation by this common denominator to get rid of the fractions: \[ (x)(x-3) - (x)(x+1) = 9(x+1)(x-3) \]

Step 2: Simplify and solve the equation

Expand the brackets and simplify: \( x^2 - 3x - x^2 + x = 9x^2 - 18x - 27 \). Simplifying gives \( 9x^2 - 18x - x + 3x - 27 = 0 \), which simplifies to \( 9x^2 - 16x - 27 = 0 \). Solve this quadratic equation by factorisation or using the quadratic formula.

Step 3: Solve the quadratic equation

We will use the Quadratic Formula to solve the equation \( 9x^2 - 16x - 27 = 0 \), which is \( x = \[ \frac{-b \pm \sqrt {b^2 - 4ac}}{2a} \] \), where \( a = 9 \), \( b = -16 \) and \( c = -27 \). Solve to find the values of \( x \).

Step 4: Check the solution

Substitute the found values of \( x \) back into the original equation to verify the solution. If the left-hand side equals the right-hand side, then the solution is verified.

Key Concepts

Solving Equations

The process of solving equations involves finding the value of the variable that makes the equation true. It is akin to solving a mystery, where each operation brings us closer to discovering the answer owed to a balance between both sides of an equation. Consider the equation as a balanced scale; what you do to one side, you must do to the other.
If we look at our exercise, the key element is maintaining balance while manipulating the equation to find the solution for the variable in question. Each step should clearly guide the transition from the equation to the isolated variable. An equation can be as simple as a linear equation, or more complex like the quadratic equation in our exercise.

• Start by identifying what type of equation you're dealing with.
• Look for ways to simplify or rearrange it to isolate the variable.
• Ensure that each step keeps the equation balanced and follows logical rules of arithmetic.

Here, we began by clearing the denominators to make further calculations straightforward and then worked through simplification to bring us closer to isolating the variable.

Fractional Equations

Fractional equations contain fractions with variables in the numerators or denominators. These equations often require an extra step of manipulation to eliminate the fractions for easier solving.
To solve a fractional equation, you must first find a common denominator. This step helps in eliminating the fractions entirely by multiplying each term in the equation by this common denominator.
In our example, the equation is \( \frac{x}{x+1}-\frac{x}{x-3}=9 \). The common denominator here is \( (x+1)(x-3) \). Multiplying each term by this denominator eliminates the fractions:

• For \( \frac{x}{x+1} \), multiplying by \( (x+1)(x-3) \) leaves us with \( x(x-3) \).
• For \( \frac{x}{x-3} \), multiplying by \( (x+1)(x-3) \) leaves \( x(x+1) \).

The equation becomes a polynomial equation, which is then more straightforward to solve. Remember, the elimination of fractions simplifies subsequent steps significantly.

Quadratic Formula

Quadratic equations play a significant role in algebra, defining equations of the form \( ax^2 + bx + c = 0 \). Solving these can sometimes require using the Quadratic Formula, especially when factoring isn't straightforward.
The Quadratic Formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). It provides a reliable method to find roots for any quadratic equation, whether they are simple integer solutions or complex numbers.
For example, in our exercise, the equation \( 9x^2 - 16x - 27 = 0 \) uses this formula, where \( a = 9 \), \( b = -16 \), and \( c = -27 \). Substituting these values into the formula:

• Calculate the discriminant: \( b^2 - 4ac = (-16)^2 - 4(9)(-27) \).
• Solve for \( x \) using the calculated discriminant.

The beauty of the Quadratic Formula is its universal applicability to any quadratic equation, offering a straightforward path to uncovering solutions.