Question

Solve each absolute value equation.

1. |7x+2|=10

2. |3x-5|=-1

3. |x^2-2x--16|=8

4. |3x-2|=7

5. |x^2+2x+1|=-4

6. |3x+2|+3=0

7. |3x-1|=|x+4|

8. |3/(k-1)|=4

9. |x^2-3x+3|=3

Answer 1

Solve each absolute value equation.

1. $|7x + 2| = 10$

2. $|3x - 5| = -1$

3. $|x^2 - 2x - -16| = 8$

4. $|3x - 2| = 7$

5. $|x^2 + 2x + 1| = -4$

6. $|3x + 2| + 3 = 0$

7. $|3x - 1| = |x + 4|$

8. $\left| \frac{3}{k - 1} \right| = 4$

9. $|x^2 - 3x + 3| = 3$

**Solution:**

1. $|7x + 2| = 10$

$$\left[ \begin{array}{l} 7x + 2 = 10 \\ 7x + 2 = -10 \end{array} \right] \quad \Rightarrow \quad \left[ \begin{array}{l} x = \frac{8}{7} \\ x = -\frac{12}{7} \end{array} \right] \quad \Rightarrow \quad \left[ \begin{array}{l} x = 1\frac{1}{7} \\ x = -1\frac{5}{7} \end{array} \right]$$

Answer: $1\frac{1}{7}, -1\frac{5}{7}$

2. $|3x - 5| = -1$

It has no solution, because absolute value must be only positive.

Answer: no solution

3. $|x^2 - 2x - 16| = 8$

$$\left[ \begin{array}{l} x^2 - 2x - 16 = 8 \\ x^2 - 2x - 16 = -8 \end{array} \right] \quad \Rightarrow \quad \left[ \begin{array}{l} x^2 - 2x - 24 = 0 \\ x^2 - 2x - 8 = 0 \end{array} \right] \quad \Rightarrow \quad \left[ \begin{array}{l} x_1 = 6, \quad x_2 = -4 \\ x_3 = 4, \quad x_2 = -2 \end{array} \right]$$

Answer: $-4; -2; 4; 6$

4. $|3x - 2| = 7$

$$\left[ \begin{array}{l} 3 x - 2 = 7 \\ 3 x - 2 = - 7 \end{array} \right. \quad \Rightarrow \quad \left[ \begin{array}{l} x = 3 \\ x = - \frac {5}{3} \end{array} \right. \quad \Rightarrow \quad \left[ \begin{array}{l} x = 3 \\ x = - 1 \frac {2}{3} \end{array} \right.$$

Answer: 3; -1 2/3.

5. $|x^2 + 2x + 1| = -4$

It has no solution, because absolute value must be only positive.

Answer: no solution

6. $|3x + 2| + 3 = 0$

$$| 3 x + 2 | = - 3$$

It has no solution, because absolute value must be only positive.

Answer: no solution

7. $|3x - 1| = |x + 4|$

$$\left[ \begin{array}{l} 3 x - 1 = x + 4 \\ 3 x - 1 = - x - 4 \end{array} \right. \quad \Rightarrow \quad \left[ \begin{array}{l} x = \frac {5}{2} \\ x = - \frac {3}{4} \end{array} \right. \quad \Rightarrow \quad \left[ \begin{array}{l} x = 2 \frac {1}{2} \\ x = - \frac {3}{4} \end{array} \right.$$

Answer: -3/4; 2 1/2

8. $|\frac{3}{k - 1}| = 4$

$$\left[ \begin{array}{l} \frac {3}{k - 1} = 4 \\ \frac {3}{k - 1} = - 4 \end{array} \right. \quad \Rightarrow \quad \left[ \begin{array}{l} k = \frac {3}{4} + 1 \\ k = - \frac {3}{4} + 1 \end{array} \right. \quad \Rightarrow \quad \left[ \begin{array}{l} k = 1 \frac {3}{4} \\ k = \frac {1}{4} \end{array} \right.$$

Answer: 1/4; 1 3/4

9. $|x^2 - 3x + 3| = 3$

$$\left[ \begin{array}{l} x ^ {2} - 3 x + 3 = 3 \\ x ^ {2} - 3 x + 3 = - 3 \end{array} \right. = > \left[ \begin{array}{l} x ^ {2} - 3 x = 0 \\ x ^ {2} - 3 x + 6 = 0 \end{array} \right. = > \left[ \begin{array}{l} x _ {1} = 0, \quad x _ {2} = 3 \\ D < 0 \Rightarrow \text{no solution} \end{array} \right.$$

Answer: 0; 3