Q57. - Algebra 2

Question

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

Solve each equation that you wrote in Exercise 56. Then, check each solution in the original equation, $\left|x+1\right|+2=\left|x+4\right|$. What are the solution(s) to this absolute value equation?

Step-by-Step Solution

Answer

On solving each equations value of x obtained is $-\frac{7}{2}\text{ and }\frac{3}{2}$

These values does not satisfies original equation 

So given equation has no solution.

Step 1 - Simplify x + 1 + 2 = x + 4

For simplification of given equation add the numbers then group the coefficients of variable on one side and number on other sides as shown below

$\begin{array}{c}x+1+2=x+4\\ x+3=x+4\\ x-x=4-3\\ 0=1\end{array}$

As resulting equation is never true so given equation has no solution

Step 2 - Simplify x + 1 + 2 = - x + 4

For simplification of given equation add the numbers then group the coefficients of variable on one side and number on other sides as shown below

$$\begin{gathered} x+1+2=-x-4 \\ 2x=-4-1-2 \\ x=-\frac{7}{2} \end{gathered}$$

Step 3 - Simplify - x + 1 + 2 = x + 4

To evaluate given equation, first simplify parenthesis then add the numbers then group the coefficients of variable on one side and number on other sides as shown below

$\begin{array}{c}-x-1+2=x+4\\ -x+1=x+4\\ -x-x=4-1\\ 2x=3\\ x=\frac{3}{2}\end{array}$

Step 4 - Simplify - x + 1 + 2 = - x + 4

To evaluate given equation, first simplify parenthesis then add the numbers then group the coefficients of variable on one side and number on other sides as shown below

$\begin{array}{c}-x-1+2=-x-4\\ -x+1=-x-4\\ -x+x=-4-1\\ 0=-5\end{array}$

As resulting equation is never true so given equation has no solution

Step 5 - Check each solution in original equation

To verify two obtained values of x is solution of original equation or not, replace x with these values one by one as shown below

Substitute $-\frac{7}{2}$ for x into the original equation $\left|x+1\right|+2=\left|x+4\right|$

$\begin{array}{l}|-\frac{7}{2}+1|+2\stackrel{?}{=}|-\frac{7}{2}+4|\\ \left|\frac{-5}{2}\right|+2\stackrel{?}{=}\left|\frac{1}{2}\right|\\ \frac{5}{2}+2\stackrel{?}{=}\frac{1}{2}\\ \xcancel{\frac{9}{2}=\frac{1}{2}}\end{array}$

Substitute $\frac{3}{2}$ for x into the original equation $\left|x+1\right|+2=\left|x+4\right|$

$\begin{array}{l}|\frac{3}{2}+1|+2\stackrel{?}{=}|\frac{3}{2}+4|\\ \left|\frac{5}{2}\right|+2\stackrel{?}{=}\left|\frac{11}{2}\right|\\ \frac{5}{2}+2\stackrel{?}{=}\frac{11}{2}\\ \xcancel{\frac{9}{2}=\frac{11}{2}}\end{array}$

Since both values does not satisfies original equation so original equation has no solution.