Question Find all boundary points and solve the rational inequality $\frac{x+7}{2x-3} > 1$ . Express the solution using interval notation.

Studdy Solution

STEP 1

Assumptions
1. We are given a rational inequality $\frac{x+7}{2x-3}>1$ .
2. We need to find the boundary points of the inequality.
3. We need to solve the inequality and express the solution in interval notation.
4. The inequality is not defined when the denominator is zero.

STEP 2

First, we will subtract 1 from both sides of the inequality to move all terms to one side and have zero on the other side.
$\frac{x+7}{2x-3} - 1 > 0$

STEP 3

To combine the terms into a single fraction, we need a common denominator. In this case, the common denominator is $2x-3$ .
$\frac{x+7}{2x-3} - \frac{2x-3}{2x-3} > 0$

STEP 4

Subtract the fractions by combining the numerators over the common denominator.
$\frac{(x+7) - (2x-3)}{2x-3} > 0$

STEP 5

Simplify the numerator by distributing the negative sign and combining like terms.
$\frac{x+7 - 2x + 3}{2x-3} > 0$
$\frac{-x + 10}{2x-3} > 0$

STEP 6

Now we have the inequality in the form of a single rational expression greater than zero. The next step is to find the boundary points by setting the numerator and denominator to zero separately.
$-x + 10 = 0$ $2x - 3 = 0$

STEP 7

Solve for $x$ in each equation to find the boundary points.
$-x + 10 = 0 \Rightarrow x = 10$ $2x - 3 = 0 \Rightarrow x = \frac{3}{2}$

STEP 8

The boundary points are $x = 10$ and $x = \frac{3}{2}$ . These points divide the number line into intervals that we will test in the inequality.

STEP 9

We will test the intervals created by the boundary points in the inequality $\frac{-x + 10}{2x-3} > 0$ . The intervals to test are $(-\infty, \frac{3}{2})$ , $(\frac{3}{2}, 10)$ , and $(10, \infty)$ .

STEP 10

Choose a test point from each interval and plug it into the inequality. Remember that $x = \frac{3}{2}$ and $x = 10$ are not included in the solution because they make the denominator zero and the numerator zero, respectively.

STEP 11

Test the interval $(-\infty, \frac{3}{2})$ by choosing a test point, for example, $x = 0$ .
$\frac{-0 + 10}{2(0)-3} > 0$
$\frac{10}{-3} > 0$
This is false, so the interval $(-\infty, \frac{3}{2})$ is not part of the solution.

STEP 12

Test the interval $(\frac{3}{2}, 10)$ by choosing a test point, for example, $x = 5$ .
$\frac{-5 + 10}{2(5)-3} > 0$
$\frac{5}{7} > 0$
This is true, so the interval $(\frac{3}{2}, 10)$ is part of the solution.

STEP 13

Test the interval $(10, \infty)$ by choosing a test point, for example, $x = 11$ .
$\frac{-11 + 10}{2(11)-3} > 0$
$\frac{-1}{19} > 0$
This is false, so the interval $(10, \infty)$ is not part of the solution.

STEP 14

Combine the results from the tests to write the solution in interval notation. The only interval that satisfied the inequality was $(\frac{3}{2}, 10)$ .
The solution in interval notation is:
$\left(\frac{3}{2}, 10\right)$

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