Q58. - Algebra 2

Question

For Exercises 55-58, consider the equation $\left|x+1\right|+2=\left|x+4\right|$.

58. MAKE A CONECTURE

For equations with one set of absolute value symbols, two cases must be considered. For an equation with two sets of absolute value symbols, four cases must be considered. How many cases must be considered for an equation containing three sets of absolute value symbols?

Step-by-Step Solution

Answer

8 cases must be considered for an equation containing three sets of absolute value symbols.

Step 1 - Define absolute value

For a real number a, $\left|a\right|=a$ if $a\ge 0$ and $\left|a\right|=-a$ if a<0 

Step 2 - Number of cases with third absolute value &#8805; 0

Since the number of cases with two sets of absolute value symbols is 4, now consider the third set of it as greater than or equal to 0. Since this third set will be considered in each of the 4 cases separately, so there are 4 cases with a third absolute value $\ge 0$ 

Step 3 - Number of cases with third absolute value &#60; 0

Since the number of cases with two sets of absolute value symbols is 4, now consider the third set of it as less than 0. Since this third set will be considered in each of the 4 cases separately, so there are 4 cases with a third absolute value <0.

So there is a total of 8 cases in an equation containing 3 sets of absolute values.