Section 2.11 : Linear Inequalities
7. If \(0 \le x < 3\) determine \(a\) and \(b\) for the inequality : \(a \le 4x + 1 < b\)
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This problem is really the reverse of the previous problems in this section. In the previous problems we started with something like the second inequality (of course we also had numbers in the two outer portions instead of \(a\) and \(b\)) and we had to manipulate it to get the \(x\) by itself in the middle.
The process here is basically the same just in reverse. We need to do algebraic manipulations to make the middle part of the first inequality look like the middle part of the second manipulation. The only real difference is that with the solving problems we added/subtracted the number before we dealt with the coefficient of the \(x\). Here we need to get the coefficient on the \(x\) before we get the number.
So, the first thing we’ll do is multiply all three parts of the first inequality by 4. This gives,
\[0\left( 4 \right) \le 4x < 3\left( 4 \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}0 \le 4x < 12\] Show Step 2Now all we need to do is add one to all three parts.
\[1 \le 4x + 1 < 13\] Show Step 3Comparing this inequality in the second step to the second inequality in the problem statement we can see that we must have \(\require{bbox} \bbox[2pt,border:1px solid black]{{a = 1}}\) and \(\require{bbox} \bbox[2pt,border:1px solid black]{{b = 13}}\) .