Absolute value equations and inequalities | math4u.vsb.cz

9000026409

Level:

B

Consider the following equation.

| 2 x - 4 | = 5 x - 7

Solving the equation on the intervals where it is possible to evaluate the absolute value we get equations on partial subintervals as follows.

\begin{aligned} for x & \in (- \infty; 2) : & for x & \in [2; \infty) : \\ - 2 x + 4 & = 5 x - 7 & 2 x - 4 & = 5 x - 7 \\ - 7 x & = - 11 & - 3 x & = - 3 \\ x & = \frac{11}{7} & x & = 1 \end{aligned}

Find the solution set of the original equation.

{\frac{11}{7}}

{\frac{11}{7}; 1}

{1}

\emptyset

9000027310

Level:

A

Identify the solution set of the following inequality.

| 2 x + 11 | > 0

(- \infty; - 5.5) \cup (- 5.5; \infty)

(- 2; 11)

(- \infty; - 11) \cup (11; \infty)

\emptyset

9000027308

Level:

A

Identify the solution set of the following inequality.

| 2 x - 1 | > 5

(- \infty; - 2) \cup (3; \infty)

(- \infty; - 4.5) \cup (5.5; \infty)

(1.5; \infty)

(- \infty; 0) \cup [5; \infty)

9000027301

Level:

A

Identify the solution set of the following inequality.

| x | < 3

(- 3; 3)

(0; 3)

[- 3; 3]

[0; 3]

9000027302

Level:

A

Identify the solution set of the following inequality.

| x | \geq 5

(- \infty; - 5] \cup [5; \infty)

(- 5; 5)

(- \infty; 5) \cup (5; \infty)

[- 5; \infty)

9000027303

Level:

A

Identify the solution set of the following inequality.

| x - 4 | \leq 1

[3; 5]

(- 5; - 3)

[- 5; - 3]

[- 1; 4]

9000027304

Level:

A

Identify the solution set of the following inequality.

| x - 1 | > 10

(- \infty; - 9) \cup (11; \infty)

(- \infty; 9) \cup (11; \infty)

[11; \infty)

(- \infty; 10) \cup [11; \infty)

9000027305

Level:

A

Identify the solution set of the following inequality.

| x + 2 | < 1

(- 3; - 1)

(1; 3)

[- 1; 3]

[- 2; 0]

9000027306

Level:

A

Identify the solution set of the following inequality.

| x + 3 | \geq 6

(- \infty; - 9] \cup [3; \infty)

(- \infty; 3] \cup [6; \infty)

(- \infty; - 3) \cup (9; \infty)

[- 3; 6]

9000026401

Level:

B

To solve the equation with absolute value using interval method we have to divide the domain of the equation by the zero point of the subexpression in the absolute value. Find this point.

2 x - 1 = 1 + | x |

0

\frac{1}{2}

- \frac{1}{2}

- 1

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