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Solve for
x.

-6x+14 < -28quad OR
quad9x+15 <= -12
Choose 1 answer:
(A)
x <= -3 or
x > 7
(B)
-3 <= x < 7
(C)
x < 7
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ -6 x+14<-28 \quad OR $\quad 9 x+15 \leq-12$ $\newline$ Choose $1$ answer: $\newline$ (A) $x \leq-3$ or x>7 $\newline$ (B) -3 \leq x<7 $\newline$ (C) x<7 $\newline$ (D) There are no solutions $\newline$ (E) All values of $x$ are solutions

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Is (x, y) a solution to the system of linear inequalities?

Full solution

Q. Solve for

x

.

\newline

-6 x+14<-28 \quad

OR

\quad 9 x+15 \leq-12

\newline

Choose

1

answer:

\newline

(A)

x \leq-3

or

x>7

\newline

(B)

-3 \leq x<7

\newline

(C)

x<7

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

Solving the first inequality: First, let's solve the inequality -6x + 14 < -28. $\newline$ Subtract $14$ from both sides to isolate the term with $x$ . $\newline$ -6x + 14 - 14 < -28 - 14 $\newline$ -6x < -42 $\newline$ Now, divide both sides by $-6$ . Remember that dividing by a negative number reverses the inequality sign. $\newline$ x > 7
Solving the second inequality: Next, let's solve the inequality $9x + 15 \leq -12$ . $\newline$ Subtract $15$ from both sides to isolate the term with $x$ . $\newline$ $9x + 15 - 15 \leq -12 - 15$ $\newline$ $9x \leq -27$ $\newline$ Now, divide both sides by $9$ . $\newline$ $x \leq -3$
Combining the solutions: Now we have two parts of the solution from the two inequalities: $\newline$ x > 7 from the first inequality, and $\newline$ $x \leq -3$ from the second inequality. $\newline$ Since the original problem states ＂OR＂ between the two inequalities, we combine the solutions. $\newline$ The solution set is all $x$ such that x > 7 or $x \leq -3$ .

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