Solve for x. -9x+2 > 18quad AND quad13 x+15 <= -4 Choose 1 answer: (A) x <= -(19)/(13) (B) x < -(16)/(9) (C) -(16)/(9) < x < -(19)/(13) (D) There are no solutions (E) All values of x are solutions

Solve first inequality: Solve the first inequality -9x + 2 > 18. Subtract 2 from both sides to isolate the term with x. -9x + 2 - 2 > 18 - 2 -9x > 16 Now, divide both sides by -9, remembering to reverse the inequality sign because we are dividing by a negative number. x < -16/9Solve second inequality: Solve the second inequality 13x + 15 ≤ -4. Subtract 15 from both sides to isolate the term with x. 13x + 15 - 15 ≤ -4 - 15 13x ≤ -19 Now, divide both sides by 13 to solve for x. x ≤ -19/13Combine solution sets: Combine the solutions of the two inequalities to find the solution set. The first inequality gives us x < -16/9. The second inequality gives us x ≤ -19/13. We need to find the intersection of these two sets. Since -19/13 is less than -16/9, the solution set is all x such that x is less than -16/9 and at the same time less than or equal to -19/13. This means the solution set is x ≤ -19/13.Check for overlap: Check if there is any overlap between the two solution sets. Since -19/13 is less than -16/9, all values that are less than or equal to -19/13 are also less than -16/9. Therefore, the solution set is indeed x ≤ -19/13.

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Solve for
x.

-9x+2 > 18quad AND
quad13 x+15 <= -4
Choose 1 answer:
(A)
x <= -(19)/(13)
(B)
x < -(16)/(9)
(C)
-(16)/(9) < x < -(19)/(13)
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ -9 x+2>18 \quad AND $\quad 13 x+15 \leq-4$ $\newline$ Choose $1$ answer: $\newline$ (A) $x \leq-\frac{19}{13}$ $\newline$ (B) x<-\frac{16}{9} $\newline$ (C) \( -\frac{16}{9}

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Math Problems

Algebra 1

Domain and range of quadratic functions: equations

Full solution

Q. Solve for

x

\newline

-9 x+2>18 \quad

AND

\quad 13 x+15 \leq-4

\newline

Choose

1

answer:

\newline

(A)

x \leq-\frac{19}{13}

\newline

(B)

x<-\frac{16}{9}

\newline

(C)

-\frac{16}{9}<x<-\frac{19}{13}

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

Solve first inequality: Solve the first inequality -9x + 2 > 18. $\newline$ Subtract $2$ from both sides to isolate the term with $x$ . $\newline$ -9x + 2 - 2 > 18 - 2 $\newline$ -9x > 16 $\newline$ Now, divide both sides by $-9$ , remembering to reverse the inequality sign because we are dividing by a negative number. $\newline$ x < -\frac{16}{9}
Solve second inequality: Solve the second inequality $13x + 15 \leq -4$ . $\newline$ Subtract $15$ from both sides to isolate the term with $x$ . $\newline$ $13x + 15 - 15 \leq -4 - 15$ $\newline$ $13x \leq -19$ $\newline$ Now, divide both sides by $13$ to solve for $x$ . $\newline$ $x \leq -\frac{19}{13}$
Combine solution sets: Combine the solutions of the two inequalities to find the solution set. $\newline$ The first inequality gives us x < -\frac{16}{9}. $\newline$ The second inequality gives us $x \leq -\frac{19}{13}$ . $\newline$ We need to find the intersection of these two sets. $\newline$ Since $-\frac{19}{13}$ is less than $-\frac{16}{9}$ , the solution set is all $x$ such that $x$ is less than $-\frac{16}{9}$ and at the same time less than or equal to $-\frac{19}{13}$ . $\newline$ This means the solution set is $x \leq -\frac{19}{13}$ .
Check for overlap: Check if there is any overlap between the two solution sets. $\newline$ Since $-\frac{19}{13}$ is less than $-\frac{16}{9}$ , all values that are less than or equal to $-\frac{19}{13}$ are also less than $-\frac{16}{9}$ . Therefore, the solution set is indeed $x \leq -\frac{19}{13}$ .