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3x < \\frac{1}{2}(4x + 4)

Question

3x < \frac{1}{2}(4x + 4)

Explanation:

Step1: Simplify the right - hand side

We use the distributive property \(a(b + c)=ab+ac\). For \(\frac{1}{2}(4x + 4)\), we have \(\frac{1}{2}\times4x+\frac{1}{2}\times4 = 2x + 2\). So the inequality becomes \(3x<2x + 2\).

Step2: Solve for \(x\)

Subtract \(2x\) from both sides of the inequality. According to the property of inequalities: if \(a < b\), then \(a - c < b - c\) (where \(c\) is a real number). Subtracting \(2x\) from both sides of \(3x<2x + 2\), we get \(3x-2x<2x + 2-2x\), which simplifies to \(x < 2\).

Answer:

\(x < 2\)