Solve for x. 5x-4 >= 12quad AND quad12 x+5 = (16)/(5) or x = -(3)/(4) (D) There are no solutions (E) All values of x are solutions

Solve first inequality: Solve the first inequality 5x - 4 ≥ 12. Add 4 to both sides to isolate the term with x on one side. 5x - 4 + 4 ≥ 12 + 4 5x ≥ 16 Now, divide both sides by 5 to solve for x. 5x/5 ≥ 16/5 x ≥ 16/5Solve second inequality: Solve the second inequality 12x + 5 ≤ -4. Subtract 5 from both sides to isolate the term with x on one side. 12x + 5 - 5 ≤ -4 - 5 12x ≤ -9 Now, divide both sides by 12 to solve for x. 12x/12 ≤ -9/12 x ≤ -3/4Combine solutions of both inequalities: Combine the solutions of both inequalities to find the common solution set. The first inequality gives us x ≥ 16/5, and the second inequality gives us x ≤ -3/4. Since there is no overlap between these two sets (no x can be both greater than or equal to 16/5 and less than or equal to -3/4 at the same time), there are no solutions that satisfy both inequalities simultaneously.

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Solve for
x.

5x-4 >= 12quad AND
quad12 x+5 <= -4
Choose 1 answer:
(A)
x >= (16)/(5) or
x <= -(3)/(4)
(B)
x <= (16)/(5)
(C)
x >= -(3)/(4)
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ $5 x-4 \geq 12 \quad$ AND $\quad 12 x+5 \leq-4$ $\newline$ Choose $1$ answer: $\newline$ (A) $x \geq \frac{16}{5}$ or $x \leq-\frac{3}{4}$ $\newline$ (B) $x \leq \frac{16}{5}$ $\newline$ (C) $x \geq-\frac{3}{4}$ $\newline$ (D) There are no solutions $\newline$ (E) All values of $x$ are solutions

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Q. Solve for

x

\newline

5 x-4 \geq 12 \quad

AND

\quad 12 x+5 \leq-4

\newline

Choose

1

answer:

\newline

(A)

x \geq \frac{16}{5}

x \leq-\frac{3}{4}

\newline

(B)

x \leq \frac{16}{5}

\newline

(C)

x \geq-\frac{3}{4}

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

Solve first inequality: Solve the first inequality $5x - 4 \geq 12$ . $\newline$ Add $4$ to both sides to isolate the term with $x$ on one side. $\newline$ $5x - 4 + 4 \geq 12 + 4$ $\newline$ $5x \geq 16$ $\newline$ Now, divide both sides by $5$ to solve for $x$ . $\newline$ $\frac{5x}{5} \geq \frac{16}{5}$ $\newline$ $x \geq \frac{16}{5}$
Solve second inequality: Solve the second inequality $12x + 5 \leq -4$ . $\newline$ Subtract $5$ from both sides to isolate the term with $x$ on one side. $\newline$ $12x + 5 - 5 \leq -4 - 5$ $\newline$ $12x \leq -9$ $\newline$ Now, divide both sides by $12$ to solve for $x$ . $\newline$ $\frac{12x}{12} \leq \frac{-9}{12}$ $\newline$ $x \leq -\frac{3}{4}$
Combine solutions of both inequalities: Combine the solutions of both inequalities to find the common solution set. $\newline$ The first inequality gives us $x \geq \frac{16}{5}$ , and the second inequality gives us $x \leq -\frac{3}{4}$ . Since there is no overlap between these two sets (no $x$ can be both greater than or equal to $\frac{16}{5}$ and less than or equal to $-\frac{3}{4}$ at the same time), there are no solutions that satisfy both inequalities simultaneously.