Question

for problems 9-11, solve each equation or inequality.

1. \\(\frac{5}{x-2} =

\frac{13}{2x-3}\\)

1. \\(3 = 2 + \frac{2}{x+2}\\)

1. \\(\frac{x+6}{x+1} < 2\\)

Explanation:

Solve Question 9

We solve the rational equation:

$$ \frac{5}{x-2} = \frac{13}{2x-3} $$

First, identify the domain restrictions:

$$ x eq 2 \quad \text{and} \quad x eq \frac{3}{2} $$

Cross-multiply to eliminate the denominators:

$$ 5(2x - 3) = 13(x - 2) $$

Distribute both sides:

$$ 10x - 15 = 13x - 26 $$

Subtract \(10x\) from both sides:

$$ -15 = 3x - 26 $$

Add \(26\) to both sides:

$$ 11 = 3x \implies x = \frac{11}{3} $$

Since \(\frac{11}{3}\) does not violate the restrictions, it is the solution.

Solve Question 10

We solve the rational equation:

$$ 3 = 2 + \frac{2}{z+2} $$

First, identify the domain restriction:

$$ z eq -2 $$

Subtract \(2\) from both sides:

$$ 1 = \frac{2}{z+2} $$

Multiply both sides by \(z+2\):

$$ z + 2 = 2 $$

Subtract \(2\) from both sides:

$$ z = 0 $$

Since \(0
eq -2\), the solution is valid.

Solve Question 11

We solve the rational inequality:

$$ \frac{x+6}{x+1} < 2 $$

First, identify the domain restriction:

$$ x eq -1 $$

Subtract \(2\) from both sides to set the inequality to \(0\):

$$ \frac{x+6}{x+1} - 2 < 0 $$

Find a common denominator:

$$ \frac{x+6 - 2(x+1)}{x+1} < 0 $$

Simplify the numerator:

$$ \frac{x+6 - 2x - 2}{x+1} < 0 \implies \frac{-x+4}{x+1} < 0 $$

Find the critical points where the numerator or denominator is zero:

$$ x = 4 \quad \text{and} \quad x = -1 $$

Test the intervals created by these critical points: \((-\infty, -1)\), \((-1, 4)\), and \((4, \infty)\).

• For \(x \in (-\infty, -1)\), let \(x = -2\):

$$ \frac{-(-2)+4}{-2+1} = \frac{6}{-1} = -6 < 0 \quad (\text{True}) $$

• For \(x \in (-1, 4)\), let \(x = 0\):

$$ \frac{-(0)+4}{0+1} = \frac{4}{1} = 4 < 0 \quad (\text{False}) $$

• For \(x \in (4, \infty)\), let \(x = 5\):

$$ \frac{-(5)+4}{5+1} = \frac{-1}{6} < 0 \quad (\text{True}) $$

Thus, the solution set is:

$$ x < -1 \quad \text{or} \quad x > 4 $$

In interval notation:

$$ (-\infty, -1) \cup (4, \infty) $$

Answer:

Question 9

\(x = \frac{11}{3}\)

Question 10

\(z = 0\)

Question 11

\(x < -1\) or \(x > 4\) (or in interval notation: \((-\infty, -1) \cup (4, \infty)\))