Question

20n + 16 ≤ 8 + 12n < 8 - 6n

Explanation:

Step1: Solve the left inequality \(20n + 16\leq8 + 12n\)

Subtract \(12n\) from both sides: \(20n-12n + 16\leq8\)
Simplify: \(8n+16\leq8\)
Subtract 16 from both sides: \(8n\leq8 - 16\)
Simplify: \(8n\leq - 8\)
Divide both sides by 8: \(n\leq - 1\)

Step2: Solve the right inequality \(8 + 12n<8 - 6n\)

Subtract 8 from both sides: \(12n< - 6n\)
Add \(6n\) to both sides: \(12n + 6n<0\)
Simplify: \(18n<0\)
Divide both sides by 18: \(n<0\)

Step3: Find the intersection of the two solutions

We have \(n\leq - 1\) from the left inequality and \(n<0\) from the right inequality. The intersection of \(n\leq - 1\) and \(n<0\) is \(n\leq - 1\)

Answer:

The solution to the compound inequality is \(n\leq - 1\)