Problem 23 - Algebra for College Students

Question

For Problems \(1-44\), solve each equation. $$ \frac{x}{x+1}-2=\frac{3}{x-3} $$

Step-by-Step Solution

Answer

No solution for x works due valid concern each showed impracticality per division or unequal value issues constraint, usually indicative impossibility original equation under typical context real checks.

Step 1: Identify denominators

Observe that the equation is \( \frac{x}{x+1} - 2 = \frac{3}{x-3} \). The denominators of the fraction terms are \( x+1 \) and \( x-3 \). We will need to address these to find a common denominator.

Step 2: Eliminate fractions by finding a common denominator

The common denominator for \( x+1 \) and \( x-3 \) is \( (x+1)(x-3) \). We multiply every term in the equation by this common denominator to eliminate the fractions:\[ (x+1)(x-3) \left( \frac{x}{x+1} - 2 \right) = (x+1)(x-3) \left( \frac{3}{x-3} \right) \]

Step 3: Simplify each term

Now, distribute the denominators over each term:- For \( \frac{x}{x+1} \), multiplying by \( (x+1)(x-3) \) results in \( x(x-3) \) because \( x+1 \) cancels.- For \(-2\), multiplying by \( (x+1)(x-3) \) gives \( -2(x+1)(x-3) \).- For \( \frac{3}{x-3} \), multiplying by \( (x+1)(x-3) \) results in \( 3(x+1) \) because \( x-3 \) cancels.Putting it together, we have: \[ x(x-3) - 2(x+1)(x-3) = 3(x+1) \]

Step 4: Expand the terms

Let's expand each part:- \( x(x-3) \) expands to \( x^2 - 3x \).- \( -2(x+1)(x-3) \) expands: - \( (x+1)(x-3) = x^2 - 3x + x - 3 = x^2 - 2x - 3 \) - So, \( -2(x^2 - 2x - 3) = -2x^2 + 4x + 6 \)- \( 3(x+1) \) expands to \( 3x + 3 \).The equation becomes:\[ x^2 - 3x - 2x^2 + 4x + 6 = 3x + 3 \]

Step 5: Combine like terms and simplify

Combine like terms from the expanded equation:- \( x^2 - 2x^2 = -x^2 \)- \(-3x + 4x = 1x \)This gives:\[ -x^2 + x + 6 = 3x + 3 \]

Step 6: Move all terms to one side

To solve for \( x \), move all terms to one side:\[ -x^2 + x + 6 - 3x - 3 = 0 \]Simplifying, we get:\[ -x^2 - 2x + 3 = 0 \]

Step 7: Multiply by -1 to simplify

Multiply the entire equation by \(-1\) to make calculations easier:\[ x^2 + 2x - 3 = 0 \]

Step 8: Solve for x using the quadratic formula

We have a quadratic equation in the form \( ax^2+bx+c=0 \), where \( a=1, b=2, c=-3 \). Using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the values:\[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)} \]Calculate discriminant and solutions:\[ x = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2} \]\[ x = \frac{-2 \pm 4}{2} \]The solutions are:- \( x = \frac{2}{2} = 1 \)- \( x = \frac{-6}{2} = -3 \)

Step 9: Verify the solutions

Check each solution in the original equation to ensure they are valid:- For \( x = 1 \): - \( \frac{1}{1+1} - 2 eq \frac{3}{1-3} \) - This is not valid.- For \( x = -3 \): - \( x + 1 = -2 \) and \( x - 3 = -6 \), these don't introduce division by zero. - Neither solution gives equal sides, but it has no division by zero so \( x = -3 \) is acceptable given provided checks might imply realistic constraint not met yet outside provided initial condition to the context.

Key Concepts

Rational Equations

A rational equation is an equation that involves one or more fractions containing polynomials in the numerator and the denominator. These types of equations can initially seem complex since they involve manipulating fractions in an algebraic context.
Understanding rational equations is important because they show up in various real-world applications, such as calculating rates, proportions, or other relationships involving multiplication and division by variables.

• To solve rational equations, you usually need to first identify the denominators of all fractions in the equation.
• The next step often involves finding a way to eliminate these fractions, typically by finding a common denominator.
• Once the fractions are eliminated, the equation can be solved like any other algebraic equation.

Remember, solving rational equations might require checking your solutions to ensure they do not make any denominator zero, as having a zero in the denominator invalidates a solution.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). This formula provides a straightforward method to find the solutions to these equations without having to factorize. It states:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here’s how you can use the quadratic formula effectively:

• Identify the coefficients \(a\), \(b\), and \(c\) from your quadratic equation.
• Substitute these values into the quadratic formula.
• Calculate the discriminant \(b^2 - 4ac\) to determine the nature of the roots:
• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is one real solution.
• If it is negative, there are no real solutions; the solutions are complex numbers.

This method is particularly useful when factorization is difficult or impossible for the given equation.

Common Denominator

When dealing with rational equations, finding a common denominator can help to eliminate the fractions, leading to a simpler equation. A common denominator is a shared multiple of the original denominators in each fraction of the equation. In the example from the step-by-step process, the fractions \(\frac{x}{x+1}\) and \(\frac{3}{x-3}\) were involved.
Here's why finding a common denominator is key:

• By multiplying every term in the equation by this common denominator, you effectively "clear" the fractions from the equation.
• This makes the equation easier to manage, turning it into a more standard algebraic problem without fractions.

The common denominator is often the least common multiple of the denominators involved. In the example, \((x+1)(x-3)\) serves as a common denominator, which both addresses the different parts of the rational equation and ensures that no unlawful division by zero occurs as long as constraints are considered.

Solving Equations

Solving equations involves finding all possible values of the variable that make the equation true. When you're working with equations, whether they're linear, quadratic, or rational, you follow similar steps:

• Start by simplifying both sides if possible.
• Get all variable terms on one side and constant terms on the other.
• Use algebra to isolate the variable.
• Check your solutions, especially in rational equations to ensure they don't make any denominator zero.

Specifically with rational equations, once the common denominator is identified and used to eliminate fractions, the resulting equation might need further simplification involving expanding brackets or combining like terms before attempting to solve for the variable. Confirming the validity of your solutions by substituting them back into the original equation is crucial to ensure no mathematical errors occurred during manipulation, especially avoiding scenarios that would include division by zero.