Inequalities I

Sol A Subtracting $\frac{x}{2}$ and 9 from both sides gives $-10<\frac{5}{2}x$, and then multiplying by $2/5$ gives $-4<x$ or $x>-4$. Therefore $(-4,\infty)$ is the solution.

Sol B Multiplying both sides out gives $x^2+4x+4>x^2+4$, and then subtracting $x^2+4$ from both sides gives $4x>0$. Therefore $x>0$, so the solution is given by $(0,\infty)$.

Sol C Solving $x+3<2x+8$ gives $-5<x$ or $x>-5$, and solving $2x+8<3x+10$ gives $-2<x$ or $x>-2$. Taking the values of $x$ satisfying both inequalities gives $x>-2$, so $(-2,\infty)$ is the solution.

Sol D Solving $2x-1\le 3x-5$ gives $4\le x$ or $x\ge 4$, and solving $3x-5\le x+9$ gives $2x\le 14$ or $x\le 7$. Taking the values of $x$ satisfying both inequalities gives $4\le x\le 7$, so $[4,7]$ is the solution.

Sol 1 $\vert 2x-5\vert\le11$ iff $-11\le2x-5\le11$ iff $-6\le2x\le16$ iff $-3\le x\le8$, so the solution is $[-3,8]$.

Sol 2 $\vert 9-2x\vert<15$ iff $-15<9-2x<15$ iff $-24<-2x<6$ iff $12>x>-3$ iff $-3<x<12$, so the solution is $(-3,12)$.

Sol 3 $x^2<9$ iff $\sqrt{x^2}<\sqrt{9}$ iff $\vert x\vert<3$ iff $-3<x<3$, so $(-3,3)$ is the solution.

Sol 4 $x^2>25$ iff $\sqrt{x^2}>\sqrt{25}$ iff $\vert x\vert>5$ iff $x<-5$ or $x>5$, so the solution is $(-\infty,-5)\cup (5,\infty)$.

Sol 5 $3\sqrt{x}-1<5$ iff $3\sqrt{x}<6$ iff $\sqrt{x}<2$ iff $x<4$ and $x\ge 0$ iff $0\le x<4$, so the solution is $[0,4)$.

Sol 6 $3x+5\vert x\vert<16$ can be broken up into the following two cases: a) If $x\ge 0$, then $3x+5x<16$ gives $8x<16$ or $x<2$; so $0\le x<2$.

b) If $x<0$, then $3x-5x<16$ gives $-2x<16$ or $x>-8$; so $-8<x<0$.

Combining the solutions for the two cases gives that $(-8,2)$ is the solution.

Sol 7 $\vert x^2-10\vert\le6$ iff $-6\le x^2-10\le 6$ iff $4\le x^2\le 16$ iff $\sqrt{4}\le \sqrt{x^2}\le \sqrt{16}$ iff $2\le \vert x\vert\le 4$ iff

$2\le x\le 4$ or $2\le -x\le 4$ iff $2\le x\le 4$ or $-2\ge x\ge -4$ iff $2\le x\le 4$ or $-4\le x\le -2$, so $[-4,-2]\cup [2,4]$ is the solution.

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