Problem 67
Question
Simplify each complex rational expression. $$\frac{\frac{3}{x-2}-\frac{4}{x+2}}{\frac{7}{x^{2}-4}}$$
Step-by-Step Solution
Verified Answer
The simplified form of the given complex rational expression is \(x = 7\).
1Step 1: Identify the least common denominator
First, find the least common denominator (LCD) of the complex expression, which is the product of the denominators of each rational expression involved in the problem. Here, the denominators are \(x-2\), \(x+2\), and \(x^{2}-4\). The term \(x^{2}-4\) can be factored into \((x-2)(x+2)\), so the LCD is \((x-2)(x+2)\).
2Step 2: Rewrite each term
Next, rewrite each term in the problem by multiplying the term by the LCD over itself. This effectively converts each term into an easier to deal with single fraction. Do this for all terms to get: \\(x-2)(x+2)\left(\frac{3}{x-2}\right)-(x-2)(x+2)\left(\frac{4}{x+2}\right)=(x-2)(x+2)\left(\frac{7}{x^{2}-4}\right)\\
3Step 3: Simplify the fractions
Now, simplify the fractions. Terms will cancel from the numerator and denominator that are same: \\(3(x+2)-4(x-2)=7\\
4Step 4: Further simplify the equation
Once again, simplify the equation by performing the distribution: \\3x+6 - 4x+8= 7 \\
5Step 5: Combine like terms
Combine like terms on left side, this results in \\-x +14 = 7\\
6Step 6: Solve for x
Solve for x by subtracting 14 from both sides of the equation to isolate x, yielding \\-x = -7. Multiply through by -1 to get x = 7 as the final result.
7Step 7: Final simplified expression
The final simplified expression can inserted back in the original problem to verify. If it checks out, then the simplification is correct.
Other exercises in this chapter
Problem 67
Find each product. $$ (x+5 y)(7 x+3 y) $$
View solution Problem 67
Simplify the radical expressions if possible. $$\sqrt[3]{32}$$
View solution Problem 67
Write each number in decimal notation without the use of exponents. $$6 \times 10^{-4}$$
View solution Problem 67
Express the distance between the given numbers using absolute value. Then find the distance by evaluating the absolute value expression. 2 and 17
View solution