Proof of inequality solution

Question

I have to solve this inequality:

$$5 ≤ 4|x − 1| + |2 − 3x|$$

and prove its solution with one (or 2 or 3) of this sentences:

$$∀x∀y |xy| = |x||y|$$

$$∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x)$$

$$∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y)$$

The solution of inequality is:

$$(-\infty, \frac{1}{7}> U <\frac{11}{7}, \infty)$$

But I have a hard time with proving the solution with the sentence. E.g. if I choose the second one I get this:

5 ≤ 4(x-1) + (2-3x) ∨ 5 ≤ - [4(x-1) + (2-3x)] <=>

5 ≤ 4x - 4 + 2 - 3x ∨ 5 ≤ -(4x-4+2-3x) <=>

5 ≤ x-2 ∨ 5 ≤ -x + 2 <=> 7 ≤ x ∨ x ≤ 3

and that is wrong.

Can someone help me out, please? Sorry for bad English, that is not my first language.

EDIT: Sentences should be applicable. I have another inequality which is already solved and it was done like this:

|3x| ≤ |2x − 1|

(x ∈ R | −1 ≤ x ≤1/5)

Sentences:

∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x) (1)

∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y) (2)

Solution: We make another sentence which we have to prove:

∀x( |3x| ≤ |2x − 1| ↔ −1 ≤ x ≤ 1/5) (3)

|3x| ≤ |2x − 1| ⇔ |3x| ≤ 2x − 1 ∨ |3x| ≤ −(2x − 1) ⇔

3x ≤ 2x − 1 ∧ −3x ≤ 2x − 1 ∨ 3x ≤ −(2x − 1) ∧ −3x ≤ −(2x − 1) ⇔

x ≤ −1 ∧ 1 ≤ 5x ∨ 5x ≤ 1 ∧ −1 ≤ x ⇔

1/5 ≤ x ≤ −1 ∨ −1 ≤ x ≤1/5⇔ −1 ≤ x ≤1/5

So we proved sentence (3)

Answer 1

Part (I) fill in this table $$ \begin{array}{c|c|c|c|c|c|c|c} x & x-1&|x-1|&4|x-1| &-3x&2-3x&|2-3x|& y = 4|x-1| +|2-3x| \\ \hline \frac{-1}{6} & &&&&&& \\ \hline 0 & &&&&&& \\ \hline \frac{1}{6} &&&&&& & \\ \hline \frac{1}{3} & &&&&&& \\ \hline \frac{1}{2} & &&&&&& \\ \hline \frac{2}{3} & &&&&&& \\ \hline \frac{5}{6} & &&&&&& \\ \hline 1 & &&&&&& \\ \hline \frac{7}{6} &&&&&& & \\ \hline \frac{4}{3} & &&&&&& \\ \hline \frac{3}{2} & &&&&&& \\ \hline \frac{5}{3} & &&&&&& \\ \hline \frac{11}{6} & &&&&&& \\ \hline 2 & &&&&&& \\ \hline \end{array} $$

Part (II) draw on graph paper, horizontal axis called $x,$ vertical axis called $y$

Answer 2

The sentences (rules) you stated will not be sufficient (applicable), because the given inequality is not a single absolute value, but a sum of two.

The standard algebraic method to solve the absolute value inequality is to divide into intervals: $$5 ≤ 4|x − 1| + |2 − 3x| \Rightarrow \\ \begin{align} 1) \ &\begin{cases}x\le \frac23\\ 5\le -4(x-1)+(2-3x) \end{cases} \Rightarrow \ \ \begin{cases}x\le \frac23\\ x\le \frac1{7}\end{cases} \Rightarrow x\in (-\infty,\frac17] \ \ \text{or} \\ 2) \ &\begin{cases}\frac23< x<1 \\ 5\le -4(x-1)-(2-3x) \end{cases}\ \Rightarrow \ \ \begin{cases}\frac23< x<1\\ x\le -3\end{cases} \Rightarrow x\in \emptyset\ \ \ \text{or} \\ 3) \ &\begin{cases}1\le x \\ 5\le 4(x-1)-(2-3x) \end{cases} \Rightarrow \ \ \begin{cases}1\le x\\ x\ge \frac{11}7\end{cases} \Rightarrow x\in [\frac{11}7,+\infty)\ \ \ \ \ \ \ \ \end{align}$$ Hence, the final solution is: $$x\in (-\infty,\frac17]\cup [\frac{11}7,+\infty).$$

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Let's try: $$5 ≤ 4|x − 1| + |2 − 3x| \iff \\ 4|x-1|\ge 5-|2-3x| \iff \\ \big[4(x-1)\ge 5-|2-3x| \big] \ \ \lor \ \ \big[-4(x-1)\ge 5-|2-3x|\big] \iff \\ \big[|2-3x|\ge 9-4x \big] \ \ \lor \ \ \big[|2-3x|\ge 1+4x\big] \iff \\ \big[2-3x\ge 9-4x \lor -(2-3x)\ge 9-4x\big] \lor \big[2-3x\ge 1+4x \lor -(2-3x)\ge 1+4x\big] \iff \\ \big[x\ge 7 \lor x\ge \frac{11}7\big] \lor \big[x\le \frac17 \lor x\le -3\big] \iff \\ \big[x\ge \frac{11}7\big] \lor \big[x\le \frac17\big] \iff \\ x\in (-\infty,\frac17]\cup [\frac{11}7,+\infty)$$