Problem 21 - Intermediate Algebra

Question

Solve equation. \(\frac{2}{x-2}+\frac{10}{x+5}=\frac{2 x}{x^{2}+3 x-10}\)

Step-by-Step Solution

Answer

The solution is \(x = 1\).

Step 1: Factor the Denominator

First, observe the right-hand side of the equation, where the denominator is a quadratic expression: \[x^2 + 3x - 10\].We need to factor this expression. Find two numbers that multiply to \(-10\) and add to \(3\), which are \(5\) and \(-2\). Hence, the factorization is:\[x^2 + 3x - 10 = (x - 2)(x + 5)\].

Step 2: Rewrite the Equation

Replace the denominator in the right-hand side:\[\frac{2}{x-2} + \frac{10}{x+5} = \frac{2x}{(x-2)(x+5)}\]

Step 3: Combine Left-hand Side Fractions

To combine \(\frac{2}{x-2}\) and \(\frac{10}{x+5}\), convert them to a common denominator, which is \((x-2)(x+5)\):\[\frac{2(x+5) + 10(x-2)}{(x-2)(x+5)} = \frac{2x + 10 + 10x - 20}{(x-2)(x+5)}\]Simplify the expression:\[\frac{12x - 10}{(x-2)(x+5)}\]

Step 4: Set Equation for Numerators

Since the denominators are the same, equate the numerators:\[12x - 10 = 2x\]

Step 5: Solve for x

Rearrange the equation by subtracting \(2x\) from both sides:\[\begin{align*}12x - 2x &= 10\10x &= 10\end{align*}\]Divide both sides by 10:\[x = 1\]

Step 6: Check for Extraneous Solutions

Verify that \(x = 1\) does not invalidate any original denominators. Plug \(x = 1\) back into the denominators:- \(x - 2 \rightarrow 1 - 2 = -1 eq 0\)- \(x + 5 \rightarrow 1 + 5 = 6 eq 0\)- \((x-2)(x+5) \rightarrow (-1)(6) = -6 eq 0\)All denominators are non-zero, so \(x = 1\) is valid.

Key Concepts

Factoring Quadratics

Factoring quadratics is a crucial skill when solving rational equations, as it helps in simplifying complex algebraic expressions. When presented with a quadratic equation like \(x^2 + 3x - 10\), the aim is to break it down into a product of simpler linear terms. To do this, look for two numbers that both:

• Multiply to give the constant term of the quadratic (in this case, -10)
• Add up to the coefficient of the middle term (3 in this example)

For \(x^2 + 3x - 10\), the pair of numbers that satisfy these conditions are 5 and -2. Therefore, the quadratic can be factored into \((x - 2)(x + 5)\). This factorization is essential as it allows us to rewrite the equation with a common denominator, facilitating the solution process.

Common Denominators

Finding a common denominator is a vital step for adding or subtracting fractions. In the context of solving rational equations, it's necessary to have the same denominator to simplify and effectively work through the equation. For example, when combining fractions such as \(\frac{2}{x-2}\) and \(\frac{10}{x+5}\), a common denominator helps unify these fractions under a single denominator, making them easier to manage.For the given equation, the common denominator is formed by multiplying the individual denominator factors, resulting in \((x-2)(x+5)\). This allows you to rewrite both fractions with the same denominator:

• \(\frac{2}{x-2}\) becomes \(\frac{2(x+5)}{(x-2)(x+5)}\)
• \(\frac{10}{x+5}\) becomes \(\frac{10(x-2)}{(x-2)(x+5)}\)

Combining these, you arrive at a single fraction \(\frac{12x - 10}{(x-2)(x+5)}\), making it much easier to handle the equation as a whole.

Extraneous Solutions

Extraneous solutions are potential answers that arise during the algebraic process but do not satisfy the original equation. This often happens when both sides of an equation are manipulated, especially when factors shared by a numerator and a denominator cancel out.When solving equations like this one, it's critical to always check potential solutions by substituting back into the original equation. For example, once you find that \(x = 1\), you must test it against all parts of the original equation to confirm validity.Check all denominators:

• \(x - 2\) yields -1, which is non-zero
• \(x + 5\) yields 6, which is non-zero
• \((x-2)(x+5)\) yields -6, which is non-zero

Thus, all these checks verify that the found solution is real and not extraneous, providing assurance that \(x = 1\) is indeed correct.

Numerator and Denominator Equality

In the process of equating rational equations with a common denominator on both sides, you can simplify the problem by directly comparing the numerators. When both sides of the equation share the same denominator, it implies that the numerators must be equal for the entire factions to be equal.Consider the equation after rewriting with a common denominator: \(\frac{12x - 10}{(x-2)(x+5)} = \frac{2x}{(x-2)(x+5)}\). Both sides have the identical denominator \((x-2)(x+5)\), so you can safely remove it from consideration and equate the numerators:\[12x - 10 = 2x\]This simplification allows you to solve for \(x\) much more straightforwardly, by treating it as a simple linear equation. Isolating \(x\) leads to finding \(x = 1\), which you then verify against the original denominators to ensure validity.