Problem 45 - 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. $$ \frac{3}{x+2}+\frac{2}{x-2}=\frac{8}{(x+2)(x-2)} $$

Step-by-Step Solution

Answer

Though we obtained \(x = 2\) as a solution, this value of x cannot be the solution since it is one of the restrictions. Therefore, there is no valid solution for the given rational equation.

Step 1: Determine Restrictions

Determine the restrictions on the variable. The restrictions on x are the values that would make any of the denominators equal to zero. So we set each denominator equal to zero and solve for x: \(x + 2 = 0\) and \(x - 2 = 0\)This results in \(x = -2\) and \(x = 2\). Therefore, the restrictions on x are -2 and 2.

Step 2: Simplify the Equation

Without violating the restrictions, simplify the equation by multiplying both sides with the common denominator, which is \((x + 2)(x - 2)\). This gives:\[3(x - 2) + 2(x + 2)= 8\]

Step 3: Further Simplification and Solving

Further simplifying the equation we get:\[3x - 6 + 2x + 4 = 8\]Then we combine like terms:\[5x - 2 = 8\]Finally, solve for \(x\):\[x = \frac{8 + 2}{5} = 2\]

Step 4: Check against the Restrictions

We check our solution against our restrictions. We found earlier that \(x = 2\) was a restriction where the denominator of the equation becomes zero. Our solution is also \(x = 2\), which is not an acceptable solution for the given equation.

Key Concepts

Rational Expressions Restrictions

When dealing with rational expressions, it's crucial to understand that not all values can be used for the variables. These forbidden numbers are called restrictions.

Restrictions occur because division by zero is undefined. In the given exercise, we have rational expressions with denominators that include variables. To find these restrictions, we set each denominator equal to zero and solve for the variable. For the exercise \(\frac{3}{x+2} + \frac{2}{x-2} = \frac{8}{(x+2)(x-2)}\), the denominators \(x+2\) and \(x-2\) are separately set to zero, leading to the restrictions \(x = -2\) and \(x = 2\). It's essential to always consider these restrictions when solving rational equations to avoid unacceptable solutions.

Common Denominator Method

Combining rational expressions often involves finding a common denominator. This method is akin to finding a common language for the different fractions to communicate effectively.

To apply the common denominator method, we look for the least common multiple of all denominators. In our equation, the common denominator is \((x+2)(x-2)\). By multiplying each term of the equation by this common denominator, we ensure that the variable terms in the denominators get eliminated. As a result, the equation becomes a simpler, non-rational algebraic equation. This technique turns a rational equation into an easier one that we can solve with basic algebra.

Determining Variable Restrictions

Determining variable restrictions is a critical step before attempting to solve a rational equation. For any value that turns a denominator into zero, the expression becomes undefined, hence these values must be excluded from the solution set.

To identify these restricted values, we solve the equations formed when setting each denominator equal to zero. Looking back at our example, the denominators are \(x+2\) and \(x-2\), leading to restrictions at \(x = -2\) and \(x = 2\). It's vital to mark these restrictions clearly as they are off-limits for our variable \(x\), meaning they cannot appear as a final solution in our equation.

Simplifying Algebraic Equations

After using the common denominator method, we obtain an algebraic equation that should be simplified. This process involves combining like terms and isolating the variable to find its value.

To simplify an algebraic equation, first, distribute any multipliers and then combine the terms with the same variable. For instance, in the step of simplification, we start with \(3(x-2) + 2(x+2) = 8\), distribute to get \(3x-6 + 2x+4\), and then combine like terms to produce \(5x-2 = 8\). Finally, we isolate \(x\) by adding \(2\) to both sides of the equation and dividing by \(5\), which normally would give us the solution. In this case, however, our 'solution' is one of the restrictions, which tells us something went awry; either the problem has no solution or there was a mistake in calculations or reasoning somewhere along the way.