# Inequalities and Absolute Value

## Self-Test:

 1)       The solution set of the inequality $x-3 < \frac{5}{x+1}$ is given by: $(-2, 4)$ $(-\infty , -2) \cup (4, \infty)$ $(-\infty, -2) \cup (-1, 4)$ $(-2, 1) \cup (4, \infty)$ All $x \neq -1$.

Hint Move everything to one side, but be careful that $(x+1)$ might be negative.

 2)       Which picture represents the interval $1\leq x < 3$? Figure generated by Wolfram|Alpha on Sep. 3, 2013. Figure generated by Wolfram|Alpha on Sep. 3, 2013. Figure generated by Wolfram|Alpha on Sep. 3, 2013. Figure generated by Wolfram|Alpha on Sep. 3, 2013. Figure generated by Wolfram|Alpha on Sep. 3, 2013.

Hint Try expressing the intervals in the picture using inequalities.

 3)       The solution set of the inequality $0< | x-1 | < r$ consists of: All points within distance $1$ of $r$. All points within distance $r$ of $1$. $(1-r, 1) \cup (1, 1+r)$. $(1-r, 1+r)$ None of the above

Hint Be careful of the double inequality. Try sketching both inequalities separately, then taking a common solution.

 4)       Which picture represents the set of points $(x,y)$ that satisfy $2x-5y>10$: Figure generated by Wolfram|Alpha on Sep. 3, 2013. line: $y=\frac{2}{5} x -2$ Figure generated by Wolfram|Alpha on Sep. 3, 2013. line: $y=\frac{2}{5} x -2$ Figure generated by Wolfram|Alpha on Sep. 3, 2013. line: $y=\frac{5}{2} x -5$ Figure generated by Wolfram|Alpha on Sep. 3, 2013. line: $y=\frac{5}{2} x -5$ None of the above

Hint Be mindful of the signs of the coefficients.

 5)       Which of the following describes precisely the values of $x$ that are between $-1$ and $5$ inclusive or are strictly greater than $9$? $(-1,5)\cup(9,+\infty)$ $(-1,5)\cup[9,+\infty)$ $[-1,5)\cup(9,+\infty)$ $(-1,5]\cup(9,+\infty)$ $[-1,5]\cup(9,+\infty)$

Hint Start by sketching the points on the number line.

 6)       The solution to $2x+3<5x+12$ is: $x<-3$ $x>-3$ $x<\frac{15}{7}$ $x<-5$ $x>-5$

Hint Isolate the terms with $x$.

 7)       If you shade the region of the plane described by $y \geq 3x + 5$, then the points $( 0 , 0 )$ and $( -1 , 1 )$ are both located in the shaded region. are both located in the non-shaded region are situated so that $( 0 , 0 )$ is located in the shaded region, but $( -1 , 1 )$ is not. are situated so that $( 0 , 0 )$ is located in the non-shaded region, but $( -1 , 1 )$ is not. can not be located based on the information given.

Hint Draw the picture.

 8)       All of the solutions to $|x-2| + 2x \leq 16$ can be described by: $2\leq x \leq 6$ and $x \geq 14$ $x \leq 2$ and $6 \leq x \leq 14$ $-6 \leq x \leq 14$ $x \leq 6$ $x \geq 14$

Hint Don't forget that an inequality with absolute value is really two inequalities.

 9)       $x^3 + x^2 - 2x > 0$ for all values of x in $(1, \infty)$ $(-\infty,2)$ $(-2,0)\cup (1,\infty)$ $(-\infty,-2)\cup (0,1)$ $(-2,1)$

Hint Begin by factoring.

 10)       $|x-1|<|x-3|$ for all values of $x$ in $(1,3)$ $(-\infty,1)\cup (3,\infty)$ $(-2,\infty)$ $(2,\infty)$ $(-\infty, 2)$

Hint Graphs will help here.