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Question
of the following inequality. |2x - 7| < 1 one point graph point 1 solidity point 2 solidity change shading
Step1: Remove absolute - value
If \(|a|\lt b\) (\(b > 0\)), then \(-b\lt a\lt b\). So for \(|2x - 7|\lt1\), we have \(- 1\lt2x - 7\lt1\).
Step2: Add 7 to all parts
Adding 7 to each part of the compound - inequality \(-1\lt2x - 7\lt1\) gives \(-1 + 7\lt2x-7 + 7\lt1 + 7\), which simplifies to \(6\lt2x\lt8\).
Step3: Divide by 2
Dividing each part of \(6\lt2x\lt8\) by 2, we get \(\frac{6}{2}\lt\frac{2x}{2}\lt\frac{8}{2}\), or \(3\lt x\lt4\).
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The solution of the inequality \(|2x - 7|\lt1\) is \(3\lt x\lt4\). On the number - line, the open circles are at \(x = 3\) and \(x = 4\), and the shading is between them.