Question

1. \\(\frac{1}{2}x - 9 < 2x\\) 33. \\(\frac{1}{3}(4x + 3) geq \frac{2}{3}x + 2\\)

Explanation:

Problem 31: Solve \(\frac{1}{2}x - 9 < 2x\)

Step 1: Subtract \(\frac{1}{2}x\) from both sides

To isolate the variable terms on one side, we subtract \(\frac{1}{2}x\) from both sides of the inequality. This gives us:
\(-9 < 2x - \frac{1}{2}x\)
Simplifying the right - hand side, \(2x-\frac{1}{2}x=\frac{4x - x}{2}=\frac{3}{2}x\), so the inequality becomes \(-9<\frac{3}{2}x\)

Step 2: Multiply both sides by \(\frac{2}{3}\)

To solve for \(x\), we multiply both sides of the inequality \(-9<\frac{3}{2}x\) by \(\frac{2}{3}\). When we multiply both sides of an inequality by a positive number, the direction of the inequality sign remains the same.
\(-9\times\frac{2}{3}Calculating \(-9\times\frac{2}{3}\), we have \(\frac{- 18}{3}=-6\). So, \(x > - 6\)

Step 1: Distribute \(\frac{1}{3}\) on the left - hand side

Using the distributive property \(a(b + c)=ab+ac\), where \(a = \frac{1}{3}\), \(b = 4x\) and \(c = 3\), we get:
\(\frac{1}{3}\times4x+\frac{1}{3}\times3\geq\frac{2}{3}x + 2\)
Simplifying, \(\frac{4}{3}x + 1\geq\frac{2}{3}x+2\)

Step 2: Subtract \(\frac{2}{3}x\) from both sides

To get the variable terms on one side, we subtract \(\frac{2}{3}x\) from both sides of the inequality:
\(\frac{4}{3}x-\frac{2}{3}x + 1\geq2\)
Simplifying the left - hand side, \(\frac{4x-2x}{3}+1=\frac{2}{3}x + 1\geq2\)

Step 3: Subtract 1 from both sides

Subtract 1 from both sides of the inequality \(\frac{2}{3}x + 1\geq2\) to isolate the term with \(x\):
\(\frac{2}{3}x\geq2 - 1\)
Simplifying the right - hand side, we have \(\frac{2}{3}x\geq1\)

Step 4: Multiply both sides by \(\frac{3}{2}\)

Multiply both sides of the inequality \(\frac{2}{3}x\geq1\) by \(\frac{3}{2}\). Since \(\frac{3}{2}\) is positive, the direction of the inequality sign remains the same.
\(x\geq1\times\frac{3}{2}\)
So, \(x\geq\frac{3}{2}\)

Answer:

\(x > - 6\)

Problem 33: Solve \(\frac{1}{3}(4x + 3)\geq\frac{2}{3}x+2\)