Problem 50 - College Algebra

Question

Exercises \(31-50\) contain equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. $$ \frac{6}{x+3}-\frac{5}{x-2}=\frac{-20}{x^{2}+x-6} $$

Step-by-Step Solution

Answer

The solution to the equation should not include the values -3 and 2 as these make the denominators zero, thereby rendering them invalid.

Step 1 - Identify Restrictions

Set the denominators equal to zero and solve for x separately. This will give the restrictions for x. \n For \(x+3 = 0\), x = -3 is a restriction. \n For \(x-2 = 0\), x = 2 is a restriction. \n For \(x^{2}+x-6 = 0\), by factoring we get \((x+3)(x-2) = 0\). Both x = -3 and x = 2 as restrictions, which were earlier determined.

Step 2 - Clear the denominators

To get rid of the denominators, multiply every term in the equation by \((x+3)(x-2)(x^{2}+x-6)\), so we have: \n 6(x-2)(x^{2}+x-6) - 5(x+3)(x^{2}+x-6) = -20(x+3)(x-2)

Step 3 - Simplify the equation

Expand and simplify the equation obtained in step 2. Combine like terms and rearrange to get the equation in the form of a quadratic equation.

Step 4 - Solve the quadratic equation

Use either factoring, completing the square, or the quadratic formula to solve the equation for x. Remember to check the answers against the restrictions identified in step 1. If any answers contain the restricted values, they are not valid solutions to the original equation.

Key Concepts

Identifying Restrictions in Rational Equations

When dealing with rational equations, one of the first tasks is to identify the values that variables cannot take, known as restrictions. These are values that would make any denominator equal to zero, which is not permissible because division by zero is undefined. To find these, we simply set each denominator equal to zero and solve for the variable.

For example, consider the equation \[\frac{6}{x+3}-\frac{5}{x-2}=\frac{-20}{x^{2}+x-6}\]. To identify the restrictions, look at the denominators \(x+3\), \(x-2\), and \(x^{2}+x-6\). Setting \(x+3 = 0\) gives a restriction of \(x = -3\). Similarly, \(x-2 = 0\) indicates a restriction of \(x = 2\). The quadratic denominator \(x^{2}+x-6\) factors to \(x+3)(x-2)\), revealing the same restrictions. Therefore, for the equation provided, the variable \(x\) cannot take the values -3 or 2.

Clearing Denominators in Equations

Clearing the denominators in a rational equation is critical for simplifying and ultimately solving the equation. This process involves multiplying both sides of the equation by a common multiple that eliminates all the denominators.

Take the equation \[\frac{6}{x+3}-\frac{5}{x-2}=\frac{-20}{x^{2}+x-6}\]. To clear the denominators, we would multiply through by the least common denominator (LCD), which in this case is the product of the distinct denominators, \(x+3)(x-2)\). By doing so, each term gets multiplied by this LCD, and as a result, the denominators are canceled out, leaving us with a polynomial equation to solve.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill in algebra. A quadratic equation is one that can be written in the standard form \(ax^2+bx+c=0\). There are various methods to solve them, including factoring, completing the square, or using the quadratic formula. After clearing the denominators in a rational equation, we often end up with a quadratic equation to solve.

For example, when you clear denominators and simplify the equation from our initial problem, the result is a quadratic equation. To solve this, you might factor it if it's factorable, or use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). Each solution must then be checked against the earlier identified restrictions to ensure it is valid. If a solution is one of the restricted values, it must be disregarded.