Question Solve the inequality $\frac{2 x+10}{x+3} \geq 1$ and find the solution set.

Studdy Solution

STEP 1

Assumptions
1. We are solving the inequality $\frac{2x+10}{x+3} \geq 1$ .
2. We need to find the set of $x$ values that satisfy this inequality.
3. We will consider the domain of the function, which excludes $x = -3$ since it would make the denominator zero.

STEP 2

Subtract 1 from both sides of the inequality to get all terms on one side and zero on the other side.
$\frac{2x+10}{x+3} - 1 \geq 0$

STEP 3

Find a common denominator to combine the terms on the left side of the inequality.
$\frac{2x+10}{x+3} - \frac{1(x+3)}{x+3} \geq 0$

STEP 4

Combine the terms over the common denominator.
$\frac{2x+10 - (x+3)}{x+3} \geq 0$

STEP 5

Distribute the negative sign in the numerator.
$\frac{2x+10 - x - 3}{x+3} \geq 0$

STEP 6

Combine like terms in the numerator.
$\frac{x+7}{x+3} \geq 0$

STEP 7

To find the solution to the inequality, we need to determine where the expression $\frac{x+7}{x+3}$ is positive, negative, or zero. We do this by finding the critical points where the numerator or denominator is zero.

STEP 8

Set the numerator equal to zero to find its critical point.
$x+7 = 0$

STEP 9

Solve for $x$ in the numerator.
$x = -7$

STEP 10

Set the denominator equal to zero to find its critical point.
$x+3 = 0$

STEP 11

Solve for $x$ in the denominator.
$x = -3$

STEP 12

Now we have two critical points: $x = -7$ and $x = -3$ . We will use these points to divide the number line into intervals and test each interval to see if the inequality holds.

STEP 13

The intervals we need to test are $(-\infty, -7)$ , $(-7, -3)$ , $(-3, \infty)$ . We cannot include $x = -3$ in the solution set because it would make the denominator zero.

STEP 14

Choose a test point from the interval $(-\infty, -7)$ , for example, $x = -8$ .

STEP 15

Plug $x = -8$ into the inequality to see if it holds.
$\frac{-8+7}{-8+3} \geq 0$

STEP 16

Simplify the expression with the test point.
$\frac{-1}{-5} \geq 0$

STEP 17

Determine the sign of the expression. Since both the numerator and the denominator are negative, the expression is positive, so the inequality holds for this interval.

STEP 18

Choose a test point from the interval $(-7, -3)$ , for example, $x = -5$ .

STEP 19

Plug $x = -5$ into the inequality to see if it holds.
$\frac{-5+7}{-5+3} \geq 0$

STEP 20

Simplify the expression with the test point.
$\frac{2}{-2} \geq 0$

STEP 21

Determine the sign of the expression. Since the numerator is positive and the denominator is negative, the expression is negative, so the inequality does not hold for this interval.

STEP 22

Choose a test point from the interval $(-3, \infty)$ , for example, $x = 0$ .

STEP 23

Plug $x = 0$ into the inequality to see if it holds.
$\frac{0+7}{0+3} \geq 0$

STEP 24

Simplify the expression with the test point.
$\frac{7}{3} \geq 0$

STEP 25

Determine the sign of the expression. Since both the numerator and the denominator are positive, the expression is positive, so the inequality holds for this interval.

STEP 26

Combine the intervals where the inequality holds. We include $x = -7$ because the inequality is non-strict (greater than or equal to) at this point, but we exclude $x = -3$ because it would make the denominator zero.

STEP 27

The solution set is $(-\infty, -7] \cup (-3, \infty)$ .
The correct answer is: $(-\infty, -7] \cup (-3, \infty)$

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