2 years ago

I need help solving this problem |5x - 1| < 1

1 answer:

4 0

Imagine what the graph would look like. Does it open upwards or downwards?

5x - 1 has a positive slope of 5 and it is reflected horizontally. The graph would be shaped like a V. A limited segment, or none, of the graph is below a certain value of y, and the rest outside it is above.

Hence, the solution must be a < x < b.

|5x - 1| < 1

5x - 1 < 1

x < 2/5

|5x - 1| < 1

5x - 1 > -1

x > 0

0 < x < 2/5

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Likurg_2 [28]

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2 years ago

I really don’t understand this and just need an explanation of how to do it.

sertanlavr [38]

$\bf ~\hspace{12em}\left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\cfrac{2n}{-3n\cdot -2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{-2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n\cdot -n}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n}\cdot \cfrac{1}{-n}$

$\bf \cfrac{1}{-3n}\cdot \boxed{1}\cdot \cfrac{1}{-n}\implies \cfrac{1}{-3n\cdot -n}\implies \cfrac{1}{3n^2}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4\implies \left( \cfrac{1}{3n^2} \right)^4\implies \stackrel{\textit{distributing the exponent}}{\cfrac{1^4}{3^4n^{2\cdot 4}}}\implies \cfrac{1}{81n^8}$

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