Solving Absolute Value Equations - High School Math

Card 0 of 6

Question

Solve for .

Answer

Divide both sides by 3.

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

Subtract 2 from both sides to solve both scenarios for .

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Question

Solve for :

Answer

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.

and

This gives us:

and

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 cannot equal a negative number.

Our final solution is then

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Question

Answer

Notice that the equation has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

Simplifying and solving this equation for gives the answer:

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Question

What are the possible values for ?

Answer

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.

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Question

Answer

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Question

Solve:

Answer

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

The equation to solve becomes .

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