Inequalities, Intervals: Problems with Solutions

Solution:
[tex]-5[/tex] is not a solution, because the inequality [tex]-5< x[/tex] is strict, so the interval is open at [tex]-5[/tex].
[tex]3[/tex] is a solution, because [tex]x\leq 3[/tex] allows equality, so the interval is closed at [tex]3[/tex].
Answer: [tex]x\in\left(-5,3\right][/tex]
Graph on the number line:

Solution:
The interval [tex]\left[7,9\right][/tex] contains the numbers [tex]x[/tex] with [tex]7\leq x\leq 9[/tex]. We check every element:
[tex]7\in\left[7,9\right][/tex]
[tex]\frac{25}{3}\approx 8.33\in\left[7,9\right][/tex]
[tex]8\in\left[7,9\right][/tex]
[tex]\frac{41}{5}=8.2\in\left[7,9\right][/tex]
[tex]9\in\left[7,9\right][/tex]
The other sets contain [tex]0[/tex] or [tex]10[/tex], which are not in [tex]\left[7,9\right][/tex].
Answer: [tex]\left\{7,\frac{25}{3},8,\frac{41}{5},9\right\}[/tex]

Solution:
We isolate [tex]x[/tex] on the left side.
[tex]2x+3-3>-2-3[/tex]
[tex]2x>-5[/tex]
Multiply both sides by [tex]\frac{1}{2}[/tex]. It is positive, so the inequality sign stays the same:
[tex]x>-\frac{5}{2}[/tex]
Answer: [tex]x\in\left(-\frac{5}{2},+\infty\right)[/tex]
Graph on the number line:

Solution:
We isolate [tex]x[/tex] on the left side.
[tex]3x-9+9<6+9[/tex]
[tex]3x<15[/tex]
Multiply both sides by [tex]\frac{1}{3}[/tex]. It is positive, so the inequality sign stays the same:
[tex]x<5[/tex]
Answer: [tex]x\in\left(-\infty,5\right)[/tex]
Graph on the number line:

Solution:
We isolate [tex]x[/tex] on the left side.
[tex]\frac{3}{2}x+4-4\leq 10-4[/tex]
[tex]\frac{3}{2}x\leq 6[/tex]
Multiply both sides by [tex]\frac{2}{3}[/tex]. It is positive, so the inequality sign stays the same:
[tex]x\leq 4[/tex]
Answer: [tex]x\in\left(-\infty,4\right][/tex]
Graph on the number line:

Solution:
We isolate [tex]x[/tex] on the left side.
[tex]5-\frac{5}{2}x-5\geq -4-5[/tex]
[tex]-\frac{5}{2}x\geq -9[/tex]
Multiply both sides by [tex]-\frac{2}{5}[/tex]. It is negative, so the inequality sign is reversed:
[tex]x\leq \frac{18}{5}[/tex]
Answer: [tex]x\in\left(-\infty,\frac{18}{5}\right][/tex]
Graph on the number line:

Solution:
We isolate [tex]x[/tex] on the left side.
[tex]\frac{5}{2}>x+x[/tex]
[tex]\frac{5}{2}>2x[/tex]
Multiply both sides by [tex]\frac{1}{2}[/tex]. It is positive, so the inequality sign stays the same:
[tex]\frac{5}{4}>x[/tex]
[tex]x<\frac{5}{4}[/tex]
Answer: [tex]x\in\left(-\infty,\frac{5}{4}\right)[/tex]
Graph on the number line:

Solution:
[tex]-1+x\geq 2x-1[/tex]
We isolate [tex]x[/tex] on the left side.
[tex]x-2x\geq -1+1[/tex]
[tex]-x\geq 0[/tex]
Multiply both sides by [tex]-1[/tex]. It is negative, so the inequality sign is reversed:
[tex]x\leq 0[/tex]
Answer: [tex]x\in\left(-\infty,0\right][/tex]
Graph on the number line:

Solution:
[tex]2+x\geq 3x-3[/tex]
We isolate [tex]x[/tex] on the left side.
[tex]x-3x\geq -3-2[/tex]
[tex]-2x\geq -5[/tex]
Multiply both sides by [tex]-\frac{1}{2}[/tex]. It is negative, so the inequality sign is reversed:
[tex]x\leq \frac{5}{2}[/tex]
Answer: [tex]x\in\left(-\infty,\frac{5}{2}\right][/tex]
Graph on the number line:

Solution:
We isolate [tex]x[/tex] on the left side.
[tex]-7x+x\leq 4-3[/tex]
[tex]-6x\leq 1[/tex]
Multiply both sides by [tex]-\frac{1}{6}[/tex]. It is negative, so the inequality sign is reversed:
[tex]x\geq -\frac{1}{6}[/tex]
Answer: [tex]x\in\left[-\frac{1}{6},+\infty\right)[/tex]
Graph on the number line:

Solution:
Multiply all three parts by [tex]-1[/tex]. It is negative, so both inequality signs are reversed:
[tex]7\geq x>-2[/tex]
[tex]-2< x\leq 7[/tex]
Answer: [tex]x\in\left(-2,7\right][/tex]
Graph on the number line:

Solution:
This means two inequalities must hold at the same time; the solution is the intersection of their solution sets:
[tex]\left\{\begin{array}{l}-\frac{20}{3}<\frac{2}{3}x\\ \frac{2}{3}x< x\end{array}\right.[/tex]
[tex]\left\{\begin{array}{l}\frac{2}{3}x>-\frac{20}{3}\\ -\frac{1}{3}x<0\end{array}\right.[/tex]
[tex]\left\{\begin{array}{l}x>-10\\ x>0\end{array}\right.[/tex]
(In the first inequality we multiply by [tex]\frac{3}{2}>0[/tex]; in the second by [tex]-3<0[/tex], which reverses the sign.)
[tex]x\in\left(-10,+\infty\right)\cap\left(0,+\infty\right)[/tex]
Answer: [tex]x\in\left(0,+\infty\right)[/tex]
Graph on the number line:

Solution:
We perform the same operation on all three parts:
[tex]-7+2< x-2+2<1+2[/tex]
[tex]-5< x<3[/tex]
Answer: [tex]x\in\left(-5,3\right)[/tex]
Graph on the number line:

Solution:
We perform the same operation on all three parts:
[tex]3+4< x-4+4\leq 10+4[/tex]
[tex]7< x\leq 14[/tex]
Answer: [tex]x\in\left(7,14\right][/tex]
Graph on the number line:

Solution:
Multiply all three parts by [tex]4[/tex] (a positive number, so the signs stay the same):
[tex]-4< x-4\leq 2[/tex]
We perform the same operation on all three parts:
[tex]-4+4< x-4+4\leq 2+4[/tex]
[tex]0< x\leq 6[/tex]
Answer: [tex]x\in\left(0,6\right][/tex]
Graph on the number line:

Solution:
Multiply all three parts by [tex]3[/tex] (a positive number, so the signs stay the same):
[tex]6\leq 4x+2\leq 30[/tex]
We perform the same operation on all three parts:
[tex]6-2\leq 4x+2-2\leq 30-2[/tex]
[tex]4\leq 4x\leq 28[/tex]
Multiply all three parts by [tex]\frac{1}{4}[/tex] (a positive number, so the signs stay the same):
[tex]1\leq x\leq 7[/tex]
Answer: [tex]x\in\left[1,7\right][/tex]
Graph on the number line:

Solution:
For [tex]B>0[/tex] we use the property [tex]|A|\leq B \iff -B\leq A\leq B[/tex]:
[tex]-9\leq x-4\leq 9[/tex]
We perform the same operation on all three parts:
[tex]-9+4\leq x-4+4\leq 9+4[/tex]
[tex]-5\leq x\leq 13[/tex]
Answer: [tex]x\in\left[-5,13\right][/tex]

Solution:
For [tex]B>0[/tex] we use the property [tex]|A|\leq B \iff -B\leq A\leq B[/tex]:
[tex]-1\leq 2x-7\leq 1[/tex]
We perform the same operation on all three parts:
[tex]-1+7\leq 2x-7+7\leq 1+7[/tex]
[tex]6\leq 2x\leq 8[/tex]
Multiply all three parts by [tex]\frac{1}{2}[/tex] (a positive number, so the signs stay the same):
[tex]3\leq x\leq 4[/tex]
Answer: [tex]x\in\left[3,4\right][/tex]

Solution:
For [tex]B>0[/tex] we use the property [tex]|A|\geq B \iff A\geq B \text{ or } A\leq -B[/tex]:
[tex]x+\sqrt{2}\geq 1 \text{ or } x+\sqrt{2}\leq -1[/tex]
[tex]x\geq 1-\sqrt{2} \text{ or } x\leq -1-\sqrt{2}[/tex]
The solution is the union of the two intervals:
Answer: [tex]x\in\left(-\infty,-1-\sqrt{2}\right]\cup\left[1-\sqrt{2},+\infty\right)[/tex]

Solution:
For [tex]B>0[/tex] we use the property [tex]|A|< B \iff -B< A< B[/tex]:
[tex]-2<\frac{3x-1}{-4}<2[/tex]
Multiply all three parts by [tex]-4[/tex]. It is negative, so both inequality signs are reversed:
[tex]8>3x-1>-8[/tex]
[tex]-8<3x-1<8[/tex]
We perform the same operation on all three parts:
[tex]-8+1<3x-1+1<8+1[/tex]
[tex]-7<3x<9[/tex]
Multiply all three parts by [tex]\frac{1}{3}[/tex] (a positive number, so the signs stay the same):
[tex]-\frac{7}{3}< x<3[/tex]
The inequality is strict, so the endpoints are not included.
Answer: [tex]x\in\left(-\frac{7}{3},3\right)[/tex]

Solution:
For [tex]B>0[/tex] we use the property [tex]|A|\geq B \iff A\geq B \text{ or } A\leq -B[/tex]:
[tex]\frac{3-5x}{3}\geq 5 \text{ or } \frac{3-5x}{3}\leq -5[/tex]
[tex]3-5x\geq 15 \text{ or } 3-5x\leq -15[/tex]
[tex]-5x\geq 12 \text{ or } -5x\leq -18[/tex]
Dividing by [tex]-5[/tex] reverses both signs:
[tex]x\leq -\frac{12}{5} \text{ or } x\geq \frac{18}{5}[/tex]
The solution is the union of the two intervals:
Answer: [tex]x\in\left(-\infty,-\frac{12}{5}\right]\cup\left[\frac{18}{5},+\infty\right)[/tex]