Solve. Graph each solution on a number line. 5-3c = -3 (B) c > -3 (C) c >= 3 (D) c <= -3

Subtract and Simplify: Subtract c from both sides of the inequality to start isolating the variable c. 5 - 3c - c = 12 / -4 Simplify the inequality. c >= -3Graph the Solution: Graph the solution on a number line. The inequality c >= -3 means that c is greater than or equal to -3. This includes -3 and all numbers to the right of -3 on the number line. We use a closed circle at -3 to indicate that -3 is included in the solution set.

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Algebra 2

Multiply and divide rational expressions

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Q. Solve. Graph each solution on a number line.

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5-3 c \leq c+17

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(A)

c \geq-3

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(B)

c>-3

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(C)

c \geq 3

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(D)

c \leq-3

Subtract and Simplify: Subtract $c$ from both sides of the inequality to start isolating the variable $c$ . $5 - 3c - c \leq c + 17 - c$ Simplify the inequality. $5 - 4c \leq 17$
Further Isolate $c$ : Subtract $5$ from both sides of the inequality to further isolate the variable $c$ .
$5 - 5 - 4c \leq 17 - 5$
Simplify the inequality.
$-4c \leq 12$
Divide and Solve: Divide both sides of the inequality by $-4$ to solve for $c$ . Remember that dividing by a negative number reverses the inequality sign. $\newline$ $-4c / -4 \geq 12 / -4$ $\newline$ Simplify the inequality. $\newline$ $c \geq -3$
Graph the Solution: Graph the solution on a number line. The inequality $c \geq -3$ means that $c$ is greater than or equal to $-3$ . This includes $-3$ and all numbers to the right of $-3$ on the number line. We use a closed circle at $-3$ to indicate that $-3$ is included in the solution set.