Solve for x. -7x+1 >= 22quad" OR "quad-10 x+41 >= 81 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

Solving the first inequality: First, let's solve the inequality -7x + 1 ≥ 22. Subtract 1 from both sides to isolate the term with x. -7x + 1 - 1 ≥ 22 - 1 -7x ≥ 21 Now, divide both sides by -7, remembering to reverse the inequality sign because we are dividing by a negative number. x ≤ -3Solving the second inequality: Next, let's solve the inequality -10x + 41 ≥ 81. Subtract 41 from both sides to isolate the term with x. -10x + 41 - 41 ≥ 81 - 41 -10x ≥ 40 Now, divide both sides by -10, again remembering to reverse the inequality sign because we are dividing by a negative number. x ≤ -4Combining the inequalities: Now we have two inequalities: x ≤ -3 from the first inequality, and x ≤ -4 from the second inequality. Since we are looking for values of x that satisfy either inequality (this is an ＂or＂ problem), we take the union of the two sets. The solution set is all x values that are less than or equal to -3, which includes all x values that are less than or equal to -4.Final solution set for x: The final solution set for x is x ≤ -3, because this includes all the values that are also less than or equal to -4.

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Solve for
x.

-7x+1 >= 22quad" OR "quad-10 x+41 >= 81
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 $x$ . $\newline$ $-7 x+1 \geq 22 \quad \text { OR } \quad-10 x+41 \geq 81$ $\newline$ 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

-7 x+1 \geq 22 \quad \text { OR } \quad-10 x+41 \geq 81

\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

Solving the first inequality: First, let's solve the inequality $-7x + 1 \geq 22$ . $\newline$ Subtract $1$ from both sides to isolate the term with $x$ . $\newline$ $-7x + 1 - 1 \geq 22 - 1$ $\newline$ $-7x \geq 21$ $\newline$ Now, divide both sides by $-7$ , remembering to reverse the inequality sign because we are dividing by a negative number. $\newline$ $x \leq -3$
Solving the second inequality: Next, let's solve the inequality $-10x + 41 \geq 81$ . $\newline$ Subtract $41$ from both sides to isolate the term with $x$ . $\newline$ $-10x + 41 - 41 \geq 81 - 41$ $\newline$ $-10x \geq 40$ $\newline$ Now, divide both sides by $-10$ , again remembering to reverse the inequality sign because we are dividing by a negative number. $\newline$ $x \leq -4$
Combining the inequalities: Now we have two inequalities: $\newline$ $x \leq -3$ from the first inequality, and $\newline$ $x \leq -4$ from the second inequality. $\newline$ Since we are looking for values of $x$ that satisfy either inequality (this is an ＂or＂ problem), we take the union of the two sets. $\newline$ The solution set is all $x$ values that are less than or equal to $-3$ , which includes all $x$ values that are less than or equal to $-4$ .
Final solution set for x: The final solution set for x is $x \leq -3$ , because this includes all the values that are also less than or equal to $-4$ .