Question

1. solve the inequalities \\(\frac{y - 2}{3} < \frac{2y + 1}{5} \leq 3\\).

Explanation:

Step1: Split the compound inequality

We have two inequalities to solve: \(\frac{y - 2}{3}<\frac{2y + 1}{5}\) and \(\frac{2y + 1}{5}\leq3\).

Step2: Solve \(\frac{y - 2}{3}<\frac{2y + 1}{5}\)

Multiply both sides by 15 (the least common multiple of 3 and 5) to eliminate denominators:
\(15\times\frac{y - 2}{3}<15\times\frac{2y + 1}{5}\)
\(5(y - 2)<3(2y + 1)\)
Expand both sides:
\(5y-10 < 6y + 3\)
Subtract \(5y\) from both sides:
\(-10 < y + 3\)
Subtract 3 from both sides:
\(y>-13\)

Step3: Solve \(\frac{2y + 1}{5}\leq3\)

Multiply both sides by 5:
\(2y + 1\leq15\)
Subtract 1 from both sides:
\(2y\leq14\)
Divide both sides by 2:
\(y\leq7\)

Step4: Combine the solutions

From Step2, we have \(y > - 13\), and from Step3, we have \(y\leq7\). So the solution to the compound inequality is the intersection of these two solution sets.

Answer:

\(-13 < y\leq7\)