Solve an absolute value inequality with graphs - High school Algebra I Exercises

Solve the following inequalities using graphs.

\left| 3x-4 \right| \gt 2

\dfrac{3}{4} \le x \le 2

x \lt \dfrac{2}{3} \,\, \text{or} \,\, x \gt 2

x \lt \dfrac{3}{4} \,\, \text{or} \,\, x \gt 2

\dfrac{2}{3} \lt x \lt 2

Consider the following graphs:

\begin{cases} y=2 \cr \cr y=|3x-4| \end{cases}

The area in which the inequality holds is above the first graph and below the second graph (the green area).

Since the graphs intersect at \left(\dfrac{2}{3},2\right) and (2,2), we can deduce that the inequality holds for x \lt \dfrac{2}{3} \,\, \text{and} \,\, x \gt 2.

x is a solution if and only if x \lt \dfrac{2}{3} \,\, \text{or} \,\, x \gt 2.

\left| x-2 \right|\le 4

x \le -2 \,\, \text{or} \,\, x \ge 6

-6 \le x\le 2

x \le -6 \,\, \text{or} \,\, x \ge 2

-2 \le x\le 6

Coonsider the following graphs:

\begin{cases} y=|x-2| \cr \cr y=4 \end{cases}

The area in which the inequality holds is above the first graph and below the second graph (the green area).

Since the graphs intersect at \left(-2{,}4\right) and (6,4), we can deduce that the inequality holds for -2 \le x\le 6.

x is a solution if and only if -2 \le x\le 6.

2-\left|x+1\right| \ge -1

-4 \le x \le 2

-2 \le x \le 4

x \le -4\,\, \text{or} \,\, x \ge 2

x \le -2 \,\, \text{or} \,\, x \ge 2

Consider the following graphs:

\begin{cases} y=-1 \cr \cr y=2-|x+1| \end{cases}

The area in which the inequality holds is above the first graph and below the second graph (the green area).

Since the graphs intersect at (-4,-1) and (2,-1), we can deduce that the inequality holds for -4 \le x \le 2.

x is a solution if and only if -4 \le x \le 2.

\left|2x-3\right|-3 \le x

x \le -6 \,\, \text{or} \,\, x \ge 0

0 \le x \le 6

x \le 0 \,\, \text{or} \,\, x \ge 6

-6 \le x \le 0

Consider the following graphs:

\begin{cases} y=|2x-3|-3 \cr \cr y=x \end{cases}

The area in which the inequality holds is above the first graph and below the second graph (the green area).

Since the graphs intersect at \left(0{,}0\right) and (6,6), we can deduce that the inequality holds for 0 \le x \le 6.

x is a solution if and only if 0 \le x \le 6.

\left|x+2\right|\ge -x

x \ge -1

x \le -1

x \le -1 or x \ge 2

-1 \le x \le 2

Consider the following graphs:

\begin{cases} y=-x \cr \cr y=|x+2| \end{cases}

The area in which the inequality holds is above the first graph and below the second graph (the green area).

Since the graphs intersect at \left(-1{,}2\right), we can deduce that the inequality holds for x \le -1.

x is a solution if and only if x \le -1.

-\left|x+1\right|+5\ge |x|

-3 \le x \le 2

-2 \le x \le 3

x \le -2 \,\, \text{or} \,\, x \ge 3

x \le -3 \,\, \text{or} \,\, x \ge 2

Consider the following graphs:

\begin{cases} y=|x| \cr \cr y=-\left|x-1\right|+5 \end{cases}

The area in which the inequality holds is above the first graph and below the second graph (the green area).

Since the graphs intersect at \left(-3{,}3\right) and (2,2), we can deduce that the inequality holds for -3\le x \le2.

x is a solution if and only if -3\le x \le2.

\left|3x+2\right|-10 \ge -|x|

x \le -2 \,\, \text{or} \,\, x \ge 3

\left[-2{,}3\right]

x \le -3 \,\, \text{or} \,\, x \ge 2

-3 \le x \le 2

Consider the following graphs:

\begin{cases} y=-|x| \cr \cr y=|3x+2|-10 \end{cases}

The area in which the inequality holds is above the first graph and below the second graph (the green area).

Since the graphs intersect at \left(-3,-3\right) and (2,-2), we can deduce that the inequality holds for x \le -3 \,\, \text{and} \,\, x \ge 2.

x is a solution if and only if x \le -3 \,\, \text{or} \,\, x \ge 2.