Question

which of the following are solutions to $|x - 1| = 5x + 2$? check all that apply. a. $x = \frac{1}{6}$ b. $x = -\frac{4}{3}$ c. $x = \frac{3}{4}$ d. $x = -\frac{1}{6}$

Explanation:

Step1: Consider two cases for absolute value

Case 1: \(x - 1\geq0\) (i.e., \(x\geq1\)), then \(|x - 1|=x - 1\). The equation becomes \(x - 1 = 5x+2\).
Solve for \(x\):
\(x-5x=2 + 1\)
\(-4x=3\)
\(x=-\frac{3}{4}\)
But we assumed \(x\geq1\), and \(-\frac{3}{4}<1\), so this solution is extraneous.

Case 2: \(x - 1<0\) (i.e., \(x<1\)), then \(|x - 1|=-(x - 1)=-x + 1\). The equation becomes \(-x + 1=5x+2\).
Solve for \(x\):
\(-x-5x=2 - 1\)
\(-6x=1\)
\(x=-\frac{1}{6}\)
Check if \(x = -\frac{1}{6}<1\), which is true.

Step2: Verify the solution

Substitute \(x = -\frac{1}{6}\) into the original equation:
Left - hand side: \(|-\frac{1}{6}-1|=|-\frac{7}{6}|=\frac{7}{6}\)
Right - hand side: \(5\times(-\frac{1}{6})+2=-\frac{5}{6}+\frac{12}{6}=\frac{7}{6}\)
Left - hand side = Right - hand side, so \(x = -\frac{1}{6}\) is a solution.

Now check other options:

• For \(x=\frac{1}{6}\):

Left - hand side: \(|\frac{1}{6}-1|=|-\frac{5}{6}|=\frac{5}{6}\)
Right - hand side: \(5\times\frac{1}{6}+2=\frac{5}{6}+\frac{12}{6}=\frac{17}{6}
eq\frac{5}{6}\)

• For \(x = -\frac{4}{3}\):

Left - hand side: \(|-\frac{4}{3}-1|=|-\frac{7}{3}|=\frac{7}{3}\)
Right - hand side: \(5\times(-\frac{4}{3})+2=-\frac{20}{3}+\frac{6}{3}=-\frac{14}{3}
eq\frac{7}{3}\)

• For \(x=\frac{3}{4}\):

Left - hand side: \(|\frac{3}{4}-1|=|-\frac{1}{4}|=\frac{1}{4}\)
Right - hand side: \(5\times\frac{3}{4}+2=\frac{15}{4}+\frac{8}{4}=\frac{23}{4}
eq\frac{1}{4}\)

Answer:

D. \(x = -\frac{1}{6}\)