Solve for x. 3x-8 = 6 Choose 1 answer: (A) x <= (31)/(3) (B) x <= 5 (C) 5 <= x <= (31)/(3) (D) There are no solutions ㄷ) All values of x are solutions

Solve First Inequality: First, let's solve the inequality 3x - 8 ≤ 23. To isolate x, we need to add 8 to both sides of the inequality. 3x - 8 + 8 ≤ 23 + 8 3x ≤ 31 Now, divide both sides by 3 to solve for x. x ≤ 31/3Solve Second Inequality: Next, let's solve the inequality -4x + 26 ≥ 6. To isolate x, we need to subtract 26 from both sides of the inequality. -4x + 26 - 26 ≥ 6 - 26 -4x ≥ -20 Now, divide both sides by -4 to solve for x. Remember that dividing by a negative number reverses the inequality sign. x ≤ 5Combine Inequalities: Now we have two inequalities from the compound inequality: x ≤ 31/3 from the first inequality, and x ≤ 5 from the second inequality. Since the original problem states ＂OR＂ between the two inequalities, we need to find the union of the solutions. The union of x ≤ 31/3 and x ≤ 5 is all x that satisfy either one or both of the inequalities. Since 5 is less than 31/3, the solution that encompasses both is x ≤ 5.

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Solve for
x.

3x-8 <= 23quad OR
quad-4x+26 >= 6
Choose 1 answer:
(A)
x <= (31)/(3)
(B)
x <= 5
(C)
5 <= x <= (31)/(3)
(D) There are no solutions
ㄷ) All values of
x are solutions

Solve for $x$ . $\newline$ $3 x-8 \leq 23 \quad$ OR $\quad-4 x+26 \geq 6$ $\newline$ Choose $1$ answer: $\newline$ (A) $x \leq \frac{31}{3}$ $\newline$ (B) $x \leq 5$ $\newline$ (C) $5 \leq x \leq \frac{31}{3}$ $\newline$ (D) There are no solutions $\newline$ ㄷ) 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

3 x-8 \leq 23 \quad

\quad-4 x+26 \geq 6

\newline

Choose

1

answer:

\newline

(A)

x \leq \frac{31}{3}

\newline

(B)

x \leq 5

\newline

(C)

5 \leq x \leq \frac{31}{3}

\newline

(D) There are no solutions

\newline

ㄷ) All values of

x

are solutions

Solve First Inequality: First, let's solve the inequality $3x - 8 \leq 23$ . To isolate $x$ , we need to add $8$ to both sides of the inequality. $3x - 8 + 8 \leq 23 + 8$ $3x \leq 31$ Now, divide both sides by $3$ to solve for $x$ . $x \leq \frac{31}{3}$
Solve Second Inequality: Next, let's solve the inequality $-4x + 26 \geq 6$ . To isolate $x$ , we need to subtract $26$ from both sides of the inequality. $-4x + 26 - 26 \geq 6 - 26$ $-4x \geq -20$ Now, divide both sides by $-4$ to solve for $x$ . Remember that dividing by a negative number reverses the inequality sign. $x \leq 5$
Combine Inequalities: Now we have two inequalities from the compound inequality: $x \leq \frac{31}{3}$ from the first inequality, and $x \leq 5$ from the second inequality. Since the original problem states ＂OR＂ between the two inequalities, we need to find the union of the solutions. The union of $x \leq \frac{31}{3}$ and $x \leq 5$ is all $x$ that satisfy either one or both of the inequalities. Since $5$ is less than $\frac{31}{3}$ , the solution that encompasses both is $x \leq 5$ .