Problem 53 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{x+2}{x+3}-1=\frac{-1}{x^{2}+2 x-3}\)

Step-by-Step Solution

Answer

The solution is \(x=2\); \(x=-3\) is extraneous.

Step 1: Simplify and find a common denominator

Start by simplifying the left-hand side. Rewrite \(1\) as \(\frac{x+3}{x+3}\) to obtain a common denominator: \(\frac{x+2}{x+3} - \frac{x+3}{x+3}\). Perform the subtraction: \(\frac{x+2-(x+3)}{x+3} = \frac{x+2-x-3}{x+3} = \frac{-1}{x+3}\).

Step 2: Set the equation

Now that the left-hand side is simplified, the equation becomes \(\frac{-1}{x+3} = \frac{-1}{x^{2}+2x-3}\).

Step 3: Identify the factored form of the denominator

Factor the quadratic in the right-hand denominator: \(x^{2}+2x-3\). To do this, find two numbers that multiply to \(-3\) and add to \(2\). These numbers are \(3\) and \(-1\), so \(x^{2}+2x-3 = (x+3)(x-1)\).

Step 4: Cross multiply and solve

Using the equation \(\frac{-1}{x+3} = \frac{-1}{(x+3)(x-1)}\), cross multiply to clear the fractions: \(-1 \cdot (x+3)(x-1) = -1 \cdot (x+3)\). Simplify to find \((x+3)(x-1) = x+3\).

Step 5: Solve the resulting equation

The equation \((x+3)(x-1) = x+3\) implies \((x+3)(x-1) - (x+3) = 0\). This can be factored as \((x+3)(x-1-1) = (x+3)(x-2) = 0\). Thus, \(x+3=0\) or \(x-2=0\). Solving these gives \(x=-3\) or \(x=2\).

Step 6: Check for extraneous solutions

Check the original equation for \(x=-3\) and \(x=2\). Note that \(x=-3\) will cause division by zero in both the original equation fractions, so it is extraneous. Substitute \(x=2\) in the original equation: \(\frac{2+2}{2+3} - 1 = \frac{-1}{2^2+2(2)-3} \). Simplify to confirm validity: \(\frac{4}{5} - 1 = \frac{-1}{5}\). The solution holds true.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra. A quadratic equation is of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\).
The equation you're working with can have real or complex solutions. In our exercise, the equation involves finding solutions using both sides of the equation and equalizing the denominators.
Quadratic equations often appear in various formats, such as the standard form, factored form, or as part of more complex expressions like the one given. In our problem, we had to manipulate it into a standard solvable format by factoring and setting up equal fractions. This highlights why understanding how to transform and solve quadratic equations is crucial in algebra. This ability to identify quadratics hidden within expressions helps in solving a variety of problems.

Factoring

Factoring is a method of simplifying many algebraic expressions, especially quadratic equations. When factoring a quadratic, you're essentially reversing the multiplication process to break down an expression into a product of simpler terms.
In the exercise, the quadratic expression \(x^2 + 2x - 3\) was factored into \((x+3)(x-1)\).
This was crucial because it revealed the nature of the roots, or solutions, to the problem. Here's a simple way to understand and perform factoring:

• Look for two numbers that multiply to give you the constant term (-3 in this case) and add to give you the coefficient of the \(x\) term (2 here).
• Those numbers were 3 and -1, leading to our factors \((x+3)\) and \((x-1)\).

Knowing how to factor efficiently can help solve equations quickly and recognize potential extraneous solutions.

Extraneous Solutions

An extraneous solution is a solution derived from solving an equation which doesn't satisfy the original equation after substitution. This usually happens when operations such as squaring both sides introduce new solutions that aren't valid for the original equation.
In the given exercise, the original equation was manipulated, and the solution \(x=-3\) turned out to be an extraneous solution. Substituting \(x=-3\) back into the original equation leads to division by zero, clearly showing it's invalid.
To avoid extraneous solutions, it's crucial always to check all potential solutions back in the original equation.

• Ensure the solutions do not make any part of the equation undefined (like division by zero).
• Verify if the operations used to obtain the solution (e.g., squaring) do not introduce invalid results.

The valid solution \(x=2\) was checked and found not to lead to any mathematical inconsistencies, validating it as a legitimate answer.