Question

Difficulty: MediumLinear Inequalities in One Variable

If −5(2x−9)+3≥18-5(2x - 9) + 3 \geq 18, what is the maximum possible value of xx?

Answer: 3

Answer

The correct answer is 3.
Distributing the −5-5 gives −10x+45+3≥18-10x + 45 + 3 \geq 18. Combining constants yields −10x+48≥18-10x + 48 \geq 18. Subtracting 48 from both sides gives −10x≥−30-10x \geq -30. Finally, dividing both sides by −10-10 and reversing the inequality sign gives x≤3x \leq 3. The maximum possible value is therefore 3.

Step-by-Step Solution

1
Distribute −5-5 to the terms inside the parentheses.
−10x+45+3≥18-10x + 45 + 3 \geq 18
Simplify the expression by expanding the parentheses.
2
Combine the constant terms 4545 and 33 on the left side.
−10x+48≥18-10x + 48 \geq 18
Group like terms together.
3
Subtract 4848 from both sides of the inequality.
−10x≥−30-10x \geq -30
Isolate the variable term.
4
Divide both sides by −10-10 and reverse the inequality symbol.
x≤3x \leq 3
Dividing by a negative number reverses the direction of the inequality.

Key Concept

Solving multi-step linear inequalities in one variable, applying the distributive property, and reversing the inequality sign when multiplying or dividing by a negative number.
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