Question

question 1 of 2
type the correct answer in each box.
consider this expression.
\\(\frac{7}{x - 3} - \frac{5}{x + 3}\\)

find the values for \\(a\\), \\(b\\), and \\(c\\) that make this expression equivalent to the given expression.
\\(\frac{ax + b}{x^2 + c}\\)
\\(a = \square\\), \\(b = \square\\), and \\(c = \square\\)

Explanation:

Step1: Find a common denominator

The denominators are \(x - 3\) and \(x + 3\), so the common denominator is \((x - 3)(x + 3)=x^{2}-9\) (using the difference of squares formula \((a - b)(a + b)=a^{2}-b^{2}\)).
Rewrite each fraction with the common denominator:
\(\frac{7}{x - 3}=\frac{7(x + 3)}{(x - 3)(x + 3)}=\frac{7x+21}{x^{2}-9}\)
\(\frac{5}{x + 3}=\frac{5(x - 3)}{(x - 3)(x + 3)}=\frac{5x-15}{x^{2}-9}\)

Step2: Subtract the fractions

Now subtract the two fractions:
\(\frac{7x + 21}{x^{2}-9}-\frac{5x-15}{x^{2}-9}=\frac{(7x + 21)-(5x - 15)}{x^{2}-9}\)
Simplify the numerator:
\((7x+21)-(5x - 15)=7x + 21-5x + 15 = 2x+36\)
So the expression becomes \(\frac{2x + 36}{x^{2}-9}\)

Step3: Compare with \(\frac{ax + b}{x^{2}+c}\)

Compare \(\frac{2x + 36}{x^{2}-9}\) with \(\frac{ax + b}{x^{2}+c}\)
We can see that \(a = 2\), \(b = 36\), and \(c=- 9\) (since \(x^{2}+c=x^{2}-9\) implies \(c=-9\))

Answer:

\(a = 2\), \(b = 36\), \(c=-9\)