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Solve for
x.

-4x+60 < 72quad" OR "quad14 x+11 < -31
Choose 1 answer:
(A)
x < -3 or
x > -3
(B)
x > -3
(C)
x < -3
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ -4 x+60<72 \quad \text { OR } \quad 14 x+11<-31 $\newline$ Choose $1$ answer: $\newline$ (A) x<-3 or x>-3 $\newline$ (B) x>-3 $\newline$ (C) x<-3 $\newline$ (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

-4 x+60<72 \quad \text { OR } \quad 14 x+11<-31

\newline

Choose

1

answer:

\newline

(A)

x<-3

or

x>-3

\newline

(B)

x>-3

\newline

(C)

x<-3

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

Solving the first inequality: First, let's solve the inequality -4x + 60 < 72. $\newline$ Subtract $60$ from both sides to isolate the term with $x$ . $\newline$ -4x + 60 - 60 < 72 - 60 $\newline$ -4x < 12 $\newline$ Now, divide both sides by $-4$ . Remember that dividing by a negative number reverses the inequality sign. $\newline$ \frac{-4x}{-4} > \frac{12}{-4} $\newline$ x > -3
Solving the second inequality: Next, let's solve the inequality 14x + 11 < -31. $\newline$ Subtract $11$ from both sides to isolate the term with $x$ . $\newline$ 14x + 11 - 11 < -31 - 11 $\newline$ 14x < -42 $\newline$ Now, divide both sides by $14$ to solve for $x$ . $\newline$ \frac{14x}{14} < \frac{-42}{14} $\newline$ x < -3
Contradiction and no solution: Now we have two inequalities: $\newline$ x > -3 from the first inequality, and $\newline$ x < -3 from the second inequality. $\newline$ These two inequalities contradict each other, as there is no number that is both greater than and less than $-3$ at the same time. $\newline$ Therefore, there are no values of $x$ that satisfy both inequalities simultaneously.

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