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Solve for
x.

-15 x+60 <= 105quad AND
quad14 x+11 <= -31
Choose 1 answer:
(A)
x <= -3
(B)
x >= -3
(C)
x=-3
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ $-15 x+60 \leq 105 \quad$ AND $\quad 14 x+11 \leq-31$ $\newline$ Choose $1$ answer: $\newline$ (A) $x \leq-3$ $\newline$ (B) $x \geq-3$ $\newline$ (C) $x=-3$ $\newline$ (D) There are no solutions $\newline$ (E) All values of $x$ are solutions

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View step-by-step help

Solve a quadratic equation by factoring

Full solution

Q. Solve for

x

.

\newline

-15 x+60 \leq 105 \quad

AND

\quad 14 x+11 \leq-31

\newline

Choose

1

answer:

\newline

(A)

x \leq-3

\newline

(B)

x \geq-3

\newline

(C)

x=-3

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

Solve first inequality: Solve the first inequality $-15x + 60 \leq 105$ . $\newline$ Subtract $60$ from both sides to isolate the term with $x$ . $\newline$ $-15x + 60 - 60 \leq 105 - 60$ $\newline$ $-15x \leq 45$ $\newline$ Now, divide both sides by $-15$ to solve for $x$ . Remember that dividing by a negative number reverses the inequality sign. $\newline$ $-\frac{15x}{-15} \geq \frac{45}{-15}$ $\newline$ $x \geq -3$
Solve second inequality: Solve the second inequality $14x + 11 \leq -31$ . $\newline$ Subtract $11$ from both sides to isolate the term with $x$ . $\newline$ $14x + 11 - 11 \leq -31 - 11$ $\newline$ $14x \leq -42$ $\newline$ Now, divide both sides by $14$ to solve for $x$ . $\newline$ $\frac{14x}{14} \leq \frac{-42}{14}$ $\newline$ $x \leq -3$
Combine solutions: Combine the solutions from Step $1$ and Step $2$ to find the common solution set. $\newline$ From Step $1$ , we have $x \geq -3$ . $\newline$ From Step $2$ , we have $x \leq -3$ . $\newline$ The common solution that satisfies both inequalities is $x = -3$ .

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