Math / Algebra

QuestionSolve the inequality. Graph the solutions. $\left|\frac{x+6}{2}\right|>3$
Select the correct choice below and fill in any answer boxes in your choice. A. The solution is $\mathrm{x}<$ $\square$ or $x>$ $\square$ - (Type integers or simplified fractions.) B. The solution is $\square$ $<x<$ $\square$ (Type integers or simplified fractions.) C. The solution is $\mathrm{x}=$ $\square$ - (Type an integer or a simplified fraction.) D. The solution is all real numbers. E. There is no solution.

Studdy Solution

STEP 1

1. The inequality involves an absolute value function.
2. The absolute value inequality can be split into two separate inequalities.
3. The solutions of the inequalities will be solved and combined to form the final solution set.
4. The solution should be in interval notation and/or inequality notation.

STEP 2

1. Rewrite the absolute value inequality as two separate inequalities.
2. Solve each inequality separately.
3. Combine the solutions from both inequalities.
4. Graph the solutions on a number line.

STEP 3

Rewrite the absolute value inequality $\left|\frac{x+6}{2}\right| > 3$ as two separate inequalities: $\frac{x+6}{2} > 3 \quad \text{or} \quad \frac{x+6}{2} < -3$

STEP 4

Solve the first inequality $\frac{x+6}{2} > 3$ . Multiply both sides by 2: $x + 6 > 6$

STEP 5

Subtract 6 from both sides: $x > 0$

STEP 6

Solve the second inequality $\frac{x+6}{2} < -3$ . Multiply both sides by 2: $x + 6 < -6$

STEP 7

Subtract 6 from both sides: $x < -12$

STEP 8

Combine the solutions $x > 0$ and $x < -12$ . The combined solution is: $x < -12 \quad \text{or} \quad x > 0$

STEP 9

Graph the solutions on a number line: Draw two rays, one starting at $x = -12$ and extending to the left, and another starting at $x = 0$ and extending to the right.
Solution: A. The solution is $\mathrm{x} < -12$ or $x > 0$ .

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Rewrite the absolute value inequality as two separate inequalities.
2. Solve each inequality separately.
3. Combine the solutions from both inequalities.
4. Graph the solutions on a number line.

STEP 3

Rewrite the absolute value inequality $\left|\frac{x+6}{2}\right| > 3$ as two separate inequalities: $\frac{x+6}{2} > 3 \quad \text{or} \quad \frac{x+6}{2} < -3$

STEP 4

Solve the first inequality $\frac{x+6}{2} > 3$ . Multiply both sides by 2: $x + 6 > 6$

STEP 5

Subtract 6 from both sides: $x > 0$

STEP 6

Solve the second inequality $\frac{x+6}{2} < -3$ . Multiply both sides by 2: $x + 6 < -6$

STEP 7

Subtract 6 from both sides: $x < -12$

STEP 8

Combine the solutions $x > 0$ and $x < -12$ . The combined solution is: $x < -12 \quad \text{or} \quad x > 0$

STEP 9

Graph the solutions on a number line: Draw two rays, one starting at $x = -12$ and extending to the left, and another starting at $x = 0$ and extending to the right.
Solution: A. The solution is $\mathrm{x} < -12$ or $x > 0$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Rewrite the absolute value inequality as two separate inequalities.2. Solve each inequality separately.3. Combine the solutions from both inequalities.4. Graph the solutions on a number line.

STEP 3

Rewrite the absolute value inequality ∣x+62∣>3\left|\frac{x+6}{2}\right| > 3∣∣​2x+6​∣∣​>3 as two separate inequalities: x+62>3orx+62<−3 \frac{x+6}{2} > 3 \quad \text{or} \quad \frac{x+6}{2} < -3 2x+6​>3or2x+6​<−3

STEP 4

Solve the first inequality x+62>3\frac{x+6}{2} > 32x+6​>3. Multiply both sides by 2: x+6>6 x + 6 > 6 x+6>6

STEP 5

Subtract 6 from both sides: x>0 x > 0 x>0

STEP 6

Solve the second inequality x+62<−3\frac{x+6}{2} < -32x+6​<−3. Multiply both sides by 2: x+6<−6 x + 6 < -6 x+6<−6

STEP 7

Subtract 6 from both sides: x<−12 x < -12 x<−12

STEP 8

Combine the solutions x>0x > 0x>0 and x<−12x < -12x<−12. The combined solution is: x<−12orx>0 x < -12 \quad \text{or} \quad x > 0 x<−12orx>0

STEP 9

Graph the solutions on a number line: Draw two rays, one starting at x=−12x = -12x=−12 and extending to the left, and another starting at x=0x = 0x=0 and extending to the right.Solution: A. The solution is x<−12\mathrm{x} < -12x<−12 or x>0x > 0x>0.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Rewrite the absolute value inequality as two separate inequalities.
2. Solve each inequality separately.
3. Combine the solutions from both inequalities.
4. Graph the solutions on a number line.

Rewrite the absolute value inequality $\left|\frac{x+6}{2}\right| > 3$ as two separate inequalities: $\frac{x+6}{2} > 3 \quad \text{or} \quad \frac{x+6}{2} < -3$

Solve the first inequality $\frac{x+6}{2} > 3$ . Multiply both sides by 2: $x + 6 > 6$

Subtract 6 from both sides: $x > 0$

Solve the second inequality $\frac{x+6}{2} < -3$ . Multiply both sides by 2: $x + 6 < -6$

Subtract 6 from both sides: $x < -12$

Combine the solutions $x > 0$ and $x < -12$ . The combined solution is: $x < -12 \quad \text{or} \quad x > 0$

Graph the solutions on a number line: Draw two rays, one starting at $x = -12$ and extending to the left, and another starting at $x = 0$ and extending to the right.
Solution: A. The solution is $\mathrm{x} < -12$ or $x > 0$ .