91Ó°ÊÓ

When solving inequalities involving parentheses, the first step is often to use the distributive property. This property helps us eliminate the parentheses by distributing the multiplication over addition or subtraction within the parenthesis.
For instance, in the exercise \(2(u+2)-2(u-1)<5\), we need to apply the distributive property to each term inside the parentheses.
We distribute \(2\) to both \(u\) and \(2\) inside the first parenthesis, and distribute \(-2\) to each term inside the second parenthesis.
Therefore: \[2(u + 2) - 2(u - 1) < 5 \] \[2u + 4 - 2(u) - (-2) < 5 \] \[2u + 4 - 2u + 2 < 5 \]
Notice how we correctly applied -2 to each term inside the second parenthesis. This is the essence of the distributive property.

After distributing, the next essential step is to combine like terms. Like terms are terms that contain the same variable or constants. This simplifies the inequality.
For example, from our distribution step in the inequality: \[2u + 4 - 2u + 2 < 5 \]
We see that \(2u\) and \(-2u\) are like terms, and \(4\) and \(2\) are constants.
Combining the like terms: \[2u - 2u + 4 + 2 < 5 \] \[0u + 6 < 5 \] \[6 < 5 \]
Combining like terms simplifies the inequality to an easily understandable form. In this case, we end up with a simpler, but false statement.

The final step in solving the inequality is analyzing the simplified expression. This step often involves determining whether the inequality statement is true or false.
From the previous steps, we combine terms to find: \[6 < 5 \]
Clearly, 6 is not less than 5, which is a false statement. This means that no value for \(u\) will make the original inequality true.
Therefore, the solution set is empty. Analyzing the inequality helps us conclude whether there are any possible solutions to the given inequality.
If the statement were true, we would have found a range of values for \(u\) that satisfy the inequality. Here we found that no such values exist, indicating an empty solution set.