[tex]\displaystyle \sf |x-1|\leq |3x-7|[/tex]

Ao elevar ambos os lados ao quadrado garantimos que ambos os lados são positivos ( o módulo sai ). Façamos isso :

[tex]\displaystyle \sf |x-1|^2\leq |3x-7|^2\\\\ x^2-2x+1\leq 9x^2-42x+49 \\\\ x^2-2x+1-9x^2+42x-49 \leq 0 \\\\ -8x^2+40x-48\leq 0 \ \ \div 8 \\\\ -x^2+5x-6\leq 0 \\\\ -x^2+3x+2x-6 \leq 0 \\\\ -x(x-3)+2(x-3)\leq 0 \\\\ (x-3)(-x+2)\leq 0\\\\ \text{ra\'izes} : \\\\ x-3=0 \to x=3 \\\\ -x+2=0 \to x=2[/tex]

parábola com concavidade p/ baixo, positiva entre as raízes. Então x precisa estar antes da menor raiz e depois da primeira raiz, ou seja :

[tex]\displaystyle \sf \large\boxed{\sf \ S=\left\{x\geq 3\ , \ x \leq 2\right\} \ }\checkmark[/tex]

letra a

poderia também, da seguinte forma :
[tex]\displaystyle \sf |x-1|\leq |3x-7|\\\\ |x-1|-|3x-7|\leq 0 \\\\ \text{analisando os sinais das fun\c c\~oes} :\\\\ |x-1|= \left\{\begin{matrix}\sf x-1,\ se \ x \geq 1 \\\\ \sf -x+1\ , \ se \ x < 1 \end{matrix}\right \\\\\\ |3x-7|=\left\{\begin{matrix}\sf \displaystyle 3x-7,\ se \ x \geq \frac{7}{3} \\\\ \displaystyle \sf -3x+7\ , \ se \ x < \frac{7}{3} \end{matrix}\right[/tex]

[tex]\displaystyle \sf \text{subtraindo} : \\\\ |x-1|-|3x-7|= \left\{\begin{matrix}\displaystyle \sf -x+1-(-3x+7),\ se \ x < 1\\\\ \displaystyle \sf x-1-(-3x+7)\ , \ se\ 1\leq x < \frac{7}{3} \\\\ \displaystyle \sf x-1-(3x-7)\ ,\ se\ x\geq \frac{7}{3} \end{matrix}\right \\\\\\ Assim,\\\\ |x-1|-|3x-7|= \left\{\begin{matrix}\displaystyle \sf 2x-6,\ se \ x < 1\\\\ \displaystyle \sf 4x-8\ , \ se\ 1\leq x < \frac{7}{3} \\\\ \displaystyle \sf -2x+6 \ ,\ se\ x\geq \frac{7}{3} \end{matrix}\right[/tex]

agora analise o que se pede :

[tex]\displaystyle\sf |x-1|-|3x-7|\leq 0\\\\ \underline{\text{Para }x < 1} : \\\\ 2x-6\leq 0\to 2x\leq 6 \to x \leq 3\\\\ \text{ Fazendo a interse\c c\~ao com o intervalo :}\\\\ (x < 1)\cap (x\leq 3) = \boxed{\sf x < 1 } \\\\\\ \underline{\text{Para }1\leq x < \frac{7}{3}} : \\\\ 4x-8\leq 0 \to 4x\leq 8\to x \leq 2\\\\ \text{ Fazendo interse\c c\~a com o intervalo } : \\\\ \left(1\leq x < \frac{7}{3}\right) \cap(x\leq 2) = \boxed{\sf 1\leq x\leq 2\ }[/tex]

[tex]\displaystyle \sf \underline{\text{Para }x\geq \frac{7}{3}} : \\\\ -2x+6\leq 0 \to 2x \geq 6 \to x\geq 3 \\\\ \text{Fazendo a interse\c c\~ao com o intervalo} : \\\\ \left(x\geq \frac{7}{3}\right)\cap(x\geq 3) =x\geq 3 \\\\\\\ \text{Fazendo a uni\~ao de todos intervalos} : \\\\\ ( x < 1) \cup (1\leq x\leq 2) \cup (x\geq 3)= x\leq 2\ e \ x\geq 3[/tex]

letra a