Question

c (\frac{3ax + 2}{3} - \frac{5x}{2} geq -8) given that (a > \frac{5}{2})

Explanation:

Step1: Combine like terms

First, find a common denominator for the fractions, which is 6. Multiply each term by 6 to eliminate the denominators:
$$6\times\frac{3ax + 2}{3}-6\times\frac{5x}{2}\geq6\times(-8)$$
Simplify each term:
$$2(3ax + 2)-3(5x)\geq - 48$$
Expand the brackets:
$$6ax+4 - 15x\geq - 48$$

Step2: Isolate the x terms

Subtract 4 from both sides:
$$6ax-15x\geq - 48 - 4$$
Simplify the right side:
$$6ax-15x\geq - 52$$
Factor out x from the left side:
$$x(6a - 15)\geq - 52$$

Step3: Solve for x

Since \(a>\frac{5}{2}\), then \(6a-15>0\) (because if \(a = \frac{5}{2}\), \(6\times\frac{5}{2}-15=15 - 15 = 0\), and \(a>\frac{5}{2}\) means \(6a-15\) is positive). When dividing both sides of an inequality by a positive number, the inequality sign remains the same. So divide both sides by \(6a - 15\):
$$x\geq\frac{-52}{6a - 15}$$
We can simplify the denominator by factoring out 3: \(6a - 15=3(2a - 5)\), so:
$$x\geq\frac{-52}{3(2a - 5)}=\frac{-52}{6a - 15}$$ (or we can write it as \(x\geq\frac{52}{15 - 6a}\) by multiplying numerator and denominator by - 1, but both are correct)

Answer:

\(x\geq\frac{-52}{6a - 15}\) (or \(x\geq\frac{52}{15 - 6a}\))