Absolute Value

Help Questions

Math › Absolute Value

Questions 1 - 10

1

Solve for :

$\left | 2x-5\right |=3x$

Explanation

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

$\left | 2x-5\right |=3x$

and

This gives us:

and

$x=-5\ and\ 1$

However, this question has an outside of the absolute value expression, in this case . Thus, any negative value of will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, is an extraneous solution, as $\left | 2x-5\right |$ cannot equal a negative number.

Our final solution is then

2

Solve for :

$\left | 2x-5\right |=3x$

Explanation

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

$\left | 2x-5\right |=3x$

and

This gives us:

and

$x=-5\ and\ 1$

However, this question has an outside of the absolute value expression, in this case . Thus, any negative value of will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, is an extraneous solution, as $\left | 2x-5\right |$ cannot equal a negative number.

Our final solution is then

3

What are the possible values for ?

$-\sqrt{12},\sqrt{12}$

Explanation

The absolute value measures the distance from zero to the given point.

In this case, since , or , as both values are twelve units away from zero.

4

What are the possible values for ?

$-\sqrt{12},\sqrt{12}$

Explanation

The absolute value measures the distance from zero to the given point.

In this case, since , or , as both values are twelve units away from zero.

5

$\textup{What are all possible values of }b\textup{ when }\left | 3b-4\right |=11\textup{ ?}$

$5, -\frac{7}{3}$

$\frac{7}{3}, -5$

$5, -\frac{11}{3}$

$\frac{19}{3}, -5$

$5, -\frac{3}{7}$

Explanation

$\textup{Set up two equations and solve for both solutions:}$

$b = 5;;:;;;;;;;;;;;;;;;;;;;;;b=-\frac{7}{3}$

6

$\textup{What are all possible values of }b\textup{ when }\left | 3b-4\right |=11\textup{ ?}$

$5, -\frac{7}{3}$

$\frac{7}{3}, -5$

$5, -\frac{11}{3}$

$\frac{19}{3}, -5$

$5, -\frac{3}{7}$

Explanation

$\textup{Set up two equations and solve for both solutions:}$

$b = 5;;:;;;;;;;;;;;;;;;;;;;;;b=-\frac{7}{3}$

7

Solve:

$|x-5|\leqslant 0$

$-5\leqslant x\leqslant 5$

All real numbers

No solution

Explanation

The absolute value can never be negative, so the equation is ONLY valid at zero.

The equation to solve becomes .

8

Solve:

$|x-5|\leqslant 0$

$-5\leqslant x\leqslant 5$

All real numbers

No solution

Explanation

The absolute value can never be negative, so the equation is ONLY valid at zero.

The equation to solve becomes .

9

Solve for .

$3\left | x+2 \right |=18$

Explanation

$3\left | x+2 \right |=18$

Divide both sides by 3.

$\left | x+2 \right |=6$

Consider both the negative and positive values for the absolute value term.

$x+2=6\ or\ x+2=-6$

Subtract 2 from both sides to solve both scenarios for .

$x=-8\ or\ x=4$

10

Solve for .

$3\left | x+2 \right |=18$

Explanation

$3\left | x+2 \right |=18$

Divide both sides by 3.

$\left | x+2 \right |=6$

Consider both the negative and positive values for the absolute value term.

$x+2=6\ or\ x+2=-6$

Subtract 2 from both sides to solve both scenarios for .

$x=-8\ or\ x=4$

Page 1 of 2

Return to subject