Question

1. $x - \frac{7}{3+x}$
2. $\frac{x^2 - 5}{x^2 + 5x - 14} - \frac{x + 3}{x + 7}$
3. $\frac{10n}{n^2 - 25} - \frac{5}{n - 5}$
4. $\frac{6x - 3x^2}{x^2 - 5x + 6} cdot \frac{x^2 - 5x - 6}{3x + 3}$

find the vertical asymptotes.

1. $f(x) = \frac{5}{(x + 4)(3x - 1)}$
2. $f(x) = \frac{x - 1}{x^2 + 5x - 6}$

determine the vertical and horizontal asymptotes and holes if there are any.

1. $g(x) = \frac{2x + 8}{3x - 12}$
2. $f(t) = \frac{x^2 + x - 6}{x^2 - x - 12}$

1. write the equation of the graph in the format $f(x) = aleft(\frac{1}{x - h}

ight) + k$.

solve.

1. $\frac{10}{x^2 - 2x} + \frac{4}{x} = \frac{5}{x - 2}$
2. $\frac{3}{4x} + \frac{1}{8} = \frac{7}{4x}$

(image for 77: a graph with vertical asymptote $x = -3$, horizontal asymptote $y = -2$, and a point $(2, 0)$)

Explanation:

Step1: Identify the problem type

This is a rational function equation solving problem. We need to solve for \(x\) in the equation \(\frac{10}{x^{2}-2x}+\frac{4}{x}=\frac{5}{x - 2}\). First, factor the denominator of the first term: \(x^{2}-2x=x(x - 2)\). So the equation becomes \(\frac{10}{x(x - 2)}+\frac{4}{x}=\frac{5}{x - 2}\).

Step2: Find the common denominator

The common denominator for the left - hand side and the right - hand side is \(x(x - 2)\). Multiply each term by \(x(x - 2)\) to eliminate the denominators.

• For the first term \(\frac{10}{x(x - 2)}\times x(x - 2)=10\)
• For the second term \(\frac{4}{x}\times x(x - 2)=4(x - 2)\)
• For the third term \(\frac{5}{x - 2}\times x(x - 2)=5x\)

So the equation after multiplying through by \(x(x - 2)\) (where \(x
eq0\) and \(x
eq2\) to avoid division by zero) is:
\(10 + 4(x - 2)=5x\)

Step3: Expand and simplify the left - hand side

Expand \(4(x - 2)\): \(4x-8\). Then the left - hand side of the equation is \(10+4x - 8=4x + 2\). So our equation is \(4x+2 = 5x\).

Step4: Solve for \(x\)

Subtract \(4x\) from both sides of the equation \(4x+2 = 5x\):
\(4x+2-4x=5x - 4x\)
\(2=x\)

But we need to check if \(x = 2\) is a valid solution. Looking back at the original equation, when \(x = 2\), the denominators \(x^{2}-2x=2^{2}-2\times2 = 0\) and \(x - 2=0\), so \(x = 2\) makes the original equation undefined. So there is no solution.

Answer:

No solution (since \(x = 2\) is extraneous)