Quote:
nalinnair wrote
\(3r\leq{4s + 5}\)
\(|s|\leq{5}\)
Given the inequalities above, which of the following CANNOT be the value of r?
A. –20
B. –5
C. 0
D. 5
E. 20
Saumya2403 wrote:
Hi,
Please elaborate solution to this question, as above given solutions are a bit confusing
Saumya2403 , I will try.
We have to find which answer choice does not satisfy whatever solution, or range of solutions, that we find for \(r\).
The first inequality defines the solutions or ranges of solutions for \(r\) in terms of \(s\). The second inequality defines the solutions for \(s\). So we should find out what to "plug in" for \(s\) first, i.e. find out what \(s\) might be in order to plug it into the first inequality.
1) \(|s|\leq{5}\)
Remove the absolute value bars, and the expression translates to the compound expression
\(-5\leq{s}\leq{5}\)
\(s\) lies between -5 and 5, inclusive. Breaking it down further
Case One: \(s\geq {-5}\), so we will plug in -5 for \(s\) in the first inequality to test the limits of the possible solutions for \(r\)
Case Two: \(s\leq {5}\), so we will plug in 5 for \(s\)
2) Back to the first inequality: \(3r\leq{4s + 5}\)
The solutions for \(r\) depend on the solutions for \(s\) that we just found.
Case One: if \(s\geq {-5}\), then
\(3r\leq{4*(-5) + 5}\)
\(3r\leq {-15}\)
\(r\leq{-5}\)
That's one possible range of solutions for \(r\).
<------------
(-5)Case Two: if \(s\leq {5}\), then
\(3r\leq{4(5) + 5}\)
\(3r\leq{25}\)
\(r\leq{\frac{25}{3}}\), or \(r\leq{8.33}\)
That's another range of solutions for \(r\)
<-------------0-------------
8.33So the second range of solutions covers the first:
< ----(-5)----0-------------(8.33)
3) Which answer choice does not lie in that range of solutions?
A) –20: that works. -20 is less than 8.33. KEEP
B) –5: that works. -5 is less than 8.33. KEEP
C) 0: that works. 0 is less than 8.33. KEEP
D) 5: that works. 5 is less than 8. KEEP
E) 20: that DOES NOT WORK. 20 is
greater than 8.33. \(r\) must be LESS than or equal to 8.33. On the number line, 20 lies to the right of where we have defined the solutions for \(r\). 20 CANNOT be the value of \(r\). It's too large. REJECT
Answer E
Does that make sense?