Question

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if r and s can be any integers such that s > 10 and 2r + s = 15, which of the following is the solution set for r?

Explanation:

Step1: Solve for s from the equation

Given \(2r + s = 15\), we can express \(s\) in terms of \(r\) as \(s=15 - 2r\).

Step2: Apply the inequality for s

We know that \(s>10\). Substitute \(s = 15-2r\) into the inequality: \(15 - 2r>10\).

Step3: Solve the inequality for r

Subtract 15 from both sides: \(- 2r>10 - 15\), which simplifies to \(-2r>-5\).
When we divide both sides of an inequality by a negative number, the direction of the inequality sign changes. So, dividing both sides by \(- 2\) gives \(r<\frac{5}{2}\) (or \(r < 2.5\)).
Since \(r\) is an integer, the solution set for \(r\) is all integers less than \(2.5\), i.e., \(r\in\{...,-2,-1,0,1,2\}\) (all integers \(r\) where \(r < 2.5\)).

Answer:

The solution set for \(r\) is all integers \(r\) such that \(r < \frac{5}{2}\) (or in set - builder notation \(\{r\in\mathbb{Z}:r < 2.5\}\) or in list form for relevant integers \(\{...,-2,-1,0,1,2\}\)). If we assume the question had options (even though not shown here), the correct option would be the one representing all integers less than \(2.5\).