Solve for x. 12 x+7 = 40 Choose 1 answer: (A) x = (48)/(5) (B) -(3)/(2) (3)/(2) or x <= (48)/(5) (D) There are no solutions (E) All values of x are solutions

First Inequality Solution: First, we need to solve each inequality separately. Let's start with the first inequality: 12x + 7 = 40 Add 8 to both sides to isolate the term with x: 5x >= 40 + 8 5x >= 48 Now, divide both sides by 5 to solve for x: x >= 48 / 5Combined Solution: Now we have two inequalities that represent the solution set: x = 48 / 5 These two inequalities do not overlap, meaning there is no value of x that satisfies both conditions simultaneously. Therefore, there are no solutions where both inequalities are true at the same time.

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Solve for
x.

12 x+7 < -11quad AND
quad5x-8 >= 40
Choose 1 answer:
(A)
x < -(3)/(2) or
x >= (48)/(5)
(B)
-(3)/(2) < x <= (48)/(5)
(C)
x > (3)/(2) or
x <= (48)/(5)
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ 12 x+7<-11 \quad AND $\quad 5 x-8 \geq 40$ $\newline$ Choose $1$ answer: $\newline$ (A) x<-\frac{3}{2} or $x \geq \frac{48}{5}$ $\newline$ (B) $-\frac{3}{2}<x \leq="" \frac{48}{5}=""$ $\newline$ (c)="" $="" x="">\frac{3}{2}$ or $x \leq \frac{48}{5}$ $\newline$ (D) There are no solutions $\newline$ (E) All values of $x$ are solutions

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Math Problems

Precalculus

Solve rational equations

Full solution

Q. Solve for

x

\newline

12 x+7<-11 \quad

AND

\quad 5 x-8 \geq 40

\newline

Choose

1

answer:

\newline

(A)

x<-\frac{3}{2}

x \geq \frac{48}{5}

\newline

(B)

-\frac{3}{2}<x \leq \frac{48}{5}

\newline

(C)

x>\frac{3}{2}

x \leq \frac{48}{5}

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

First Inequality Solution: First, we need to solve each inequality separately. Let's start with the first inequality: $\newline$ 12x + 7 < -11 $\newline$ Subtract $7$ from both sides to isolate the term with x: $\newline$ 12x < -11 - 7 $\newline$ 12x < -18 $\newline$ Now, divide both sides by $12$ to solve for x: $\newline$ x < -\frac{18}{12} $\newline$ x < -\frac{3}{2}
Second Inequality Solution: Next, let's solve the second inequality: $\newline$ $5x - 8 \geq 40$ $\newline$ Add $8$ to both sides to isolate the term with $x$ : $\newline$ $5x \geq 40 + 8$ $\newline$ $5x \geq 48$ $\newline$ Now, divide both sides by $5$ to solve for $x$ : $\newline$ $x \geq \frac{48}{5}$
Combined Solution: Now we have two inequalities that represent the solution set: $\newline$ x < -\frac{3}{2} and $x \geq \frac{48}{5}$ $\newline$ These two inequalities do not overlap, meaning there is no value of $x$ that satisfies both conditions simultaneously. Therefore, there are no solutions where both inequalities are true at the same time.