Solve for \( x \) : CLASSWORK \( \begin{array}{ll}\text { (a) } 3 x+4<5 \\ \text { (d) } 2(x-6)>0 & \text { (2) } \\ \text { (d) } \frac{x}{3} \geq \frac{x-1}{4} & \text { (3) }\end{array} \)

Let's solve the inequalities one by one! For the first inequality \(3x + 4 < 5\): 1. Subtract 4 from both sides: \(3x < 1\) 2. Divide by 3: \(x < \frac{1}{3}\) Now, for the second inequality \(2(x-6) > 0\): 1. Expand it: \(2x - 12 > 0\) 2. Add 12 to both sides: \(2x > 12\) 3. Divide by 2: \(x > 6\) Next, let's solve \(\frac{x}{3} \geq \frac{x-1}{4}\): 1. Cross multiply to eliminate the fractions: \(4x \geq 3(x - 1)\) 2. Expand: \(4x \geq 3x - 3\) 3. Subtract \(3x\) from both sides: \(x \geq -3\) So, we have the solution sets: 1. \(x < \frac{1}{3}\) 2. \(x > 6\) 3. \(x \geq -3\) These inequalities give you a range of values for \(x\) depending on which one you're interested in!