Inequalities, Intervals: Difficult Problems with Solutions

Solution:
Let the numbers be [tex]n[/tex], [tex]n+2[/tex], [tex]n+4[/tex] ([tex]n[/tex] odd):
[tex]3n+6\le 25 \Rightarrow 3n\le 19 \Rightarrow n\le 6\frac{1}{3}[/tex]
The largest odd [tex]n[/tex] with this property is [tex]5[/tex] ([tex]5+7+9=21\le 25[/tex], while [tex]7+9+11=27>25[/tex]).

Solution:
Let the width be [tex]a[/tex]; then [tex]l=a+3[/tex]:
[tex]2(a+a+3)\le 30 \Rightarrow 4a+6\le 30 \Rightarrow a\le 6[/tex]
Also the width must be positive: [tex]a>0[/tex]. So [tex]0\lt a\le 6[/tex], and [tex]3\lt l\le 9[/tex].

Solution:
Let [tex]h[/tex] be the number of hits; then there are [tex]10-h[/tex] misses:
[tex]5h-(10-h)>30 \Rightarrow 6h>40 \Rightarrow h>6\frac{2}{3}[/tex]
The smallest whole number is [tex]h=7[/tex] ([tex]5\times 7-3=32>30[/tex]).

Solution:
[tex]x^2+x-6\le x^2-8x+16[/tex]
[tex]9x\le 22 \Rightarrow x\le 2\frac{4}{9}[/tex]
The largest integer solution is [tex]2[/tex].

Solution:
[tex](2x-1)^3=8x^3-12x^2+6x-1[/tex]
[tex]8x(x-2)(x+2)=8x(x^2-4)=8x^3-32x[/tex]
[tex]8x^3-12x^2+6x-1-8x^3+32x\lt 38x-12x^2[/tex]
[tex]-12x^2+38x-1\lt 38x-12x^2 \Rightarrow -1\lt 0[/tex]
This is true for every [tex]x[/tex], so every number is a solution.

Solution:
[tex](x-1)(x^2+x+1)=x^3-1[/tex]
[tex]x^3-1-x^3-x>1 \Rightarrow -x>2 \Rightarrow x\lt -2[/tex]
The integers in [tex]\lbrack -8, 2)[/tex] with [tex]x\lt -2[/tex] are [tex]-8[/tex], [tex]-7[/tex], [tex]-6[/tex], [tex]-5[/tex], [tex]-4[/tex], [tex]-3[/tex] – six numbers.

Solution:
a) [tex]3x+5\lt 7x-2 \Rightarrow 7\lt 4x \Rightarrow x>\frac{7}{4}[/tex]; [tex]-5x\lt -10 \Rightarrow x>2[/tex]. The solution sets are different (e.g. [tex]x=1. 8[/tex] satisfies only the first), so they are not equivalent.
b) [tex]3x-1\lt 3x+6 \Rightarrow -1\lt 6[/tex] and [tex]2x+3>2x-10 \Rightarrow 3>-10[/tex]. Both are true for every [tex]x[/tex], so they are equivalent.

Solution:
[tex]5x\lt \frac{b+2}{3} \Rightarrow x\lt \frac{b+2}{15}[/tex]
[tex]\frac{b+2}{15}=\frac{1}{4} \Rightarrow b+2=\frac{15}{4} \Rightarrow b=\frac{7}{4}=1\frac{3}{4}[/tex]

Solution:
[tex]\frac{7}{2}x-\frac{5}{2}(a-1)x\ge \frac{5}{8} \Rightarrow \frac{12-5a}{2}x\ge \frac{5}{8}[/tex]
If the coefficient of [tex]x[/tex] is not [tex]0[/tex], the inequality has solutions. If [tex]12-5a=0[/tex], it becomes [tex]0\ge \frac{5}{8}[/tex] – false for every [tex]x[/tex].
[tex]12-5a=0 \Rightarrow a=\frac{12}{5}[/tex]

Solution:
Let the order be [tex]x[/tex] shirts:
[tex]\frac{x}{20}\le \frac{x+75}{25}[/tex]
[tex]25x\le 20x+1500 \Rightarrow 5x\le 1500 \Rightarrow x\le 300[/tex]

Solution:
[tex]t[/tex] hours after Stefan's start, Emil has been riding for [tex]t+\frac{1}{3}[/tex] hours:
[tex]15t-12\left(t+\frac{1}{3}\right)>2 \Rightarrow 3t-4>2 \Rightarrow t>2[/tex]
After more than [tex]2[/tex] hours from his start.

Solution:
Let the change be [tex]k[/tex] parts per hour:
[tex]3(70+k)\ge 3(180-k) \Rightarrow 70+k\ge 180-k \Rightarrow 2k\ge 110 \Rightarrow k\ge 55[/tex]