Solve for x. -8x+3 >= 27quad" AND "quad-13 x+5 >= 57 Choose 1 answer: (A) x <= -4 (B) x <= -3 (C) -4 <= x <= -3 (D) There are no solutions (E) All values of x are solutions

Solve first inequality: First, let's solve the first inequality -8x + 3 ≥ 27. Subtract 3 from both sides to isolate the term with x. -8x + 3 - 3 ≥ 27 - 3 -8x ≥ 24 Now, divide both sides by -8. Remember that dividing by a negative number reverses the inequality sign. x ≤ -3Solve second inequality: Next, let's solve the second inequality -13x + 5 ≥ 57. Subtract 5 from both sides to isolate the term with x. -13x + 5 - 5 ≥ 57 - 5 -13x ≥ 52 Now, divide both sides by -13. Again, remember that dividing by a negative number reverses the inequality sign. x ≤ -4Find intersection of solution sets: Now we have two inequalities: x ≤ -3 x ≤ -4 To find the solution set that satisfies both inequalities, we need to take the intersection of the two sets. The intersection will be the set of all x that are less than or equal to the smaller of the two upper bounds. Since -4 is less than -3, the solution set is x ≤ -4.

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Solve for
x.

-8x+3 >= 27quad" AND "quad-13 x+5 >= 57
Choose 1 answer:
(A)
x <= -4
(B)
x <= -3
(C)
-4 <= x <= -3
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ $-8 x+3 \geq 27 \quad \text { AND } \quad-13 x+5 \geq 57$ $\newline$ Choose $1$ answer: $\newline$ (A) $x \leq-4$ $\newline$ (B) $x \leq-3$ $\newline$ (C) $-4 \leq x \leq-3$ $\newline$ (D) There are no solutions $\newline$ (E) All values of $x$ are solutions

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Math Problems

Algebra 1

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

Full solution

Q. Solve for

x

\newline

-8 x+3 \geq 27 \quad \text { AND } \quad-13 x+5 \geq 57

\newline

Choose

1

answer:

\newline

(A)

x \leq-4

\newline

(B)

x \leq-3

\newline

(C)

-4 \leq x \leq-3

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

Solve first inequality: First, let's solve the first inequality $-8x + 3 \geq 27$ . $\newline$ Subtract $3$ from both sides to isolate the term with $x$ . $\newline$ $-8x + 3 - 3 \geq 27 - 3$ $\newline$ $-8x \geq 24$ $\newline$ Now, divide both sides by $-8$ . Remember that dividing by a negative number reverses the inequality sign. $\newline$ $x \leq -3$
Solve second inequality: Next, let's solve the second inequality $-13x + 5 \geq 57$ . $\newline$ Subtract $5$ from both sides to isolate the term with $x$ . $\newline$ $-13x + 5 - 5 \geq 57 - 5$ $\newline$ $-13x \geq 52$ $\newline$ Now, divide both sides by $-13$ . Again, remember that dividing by a negative number reverses the inequality sign. $\newline$ $x \leq -4$
Find intersection of solution sets: Now we have two inequalities: $\newline$ $x \leq -3$ $\newline$ $x \leq -4$ $\newline$ To find the solution set that satisfies both inequalities, we need to take the intersection of the two sets. The intersection will be the set of all $x$ that are less than or equal to the smaller of the two upper bounds. $\newline$ Since $-4$ is less than $-3$ , the solution set is $x \leq -4$ .