What is (2x^2-1)/(5x-2)<= (4x+1)/(10) ?

The solution to (2x^2-1)/(5x-2)<= (4x+1)/(10) is 2/5 <x<= 8/3

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(2x^2-1)/(5x-2)<= (4x+1)/(10)

\frac{2 x ^{2} - 1}{5 x - 2} \leq \frac{4 x + 1}{10}

\frac{2}{5} < x \leq \frac{8}{3}

Interval Notation

(\frac{2}{5}, \frac{8}{3}]

Decimal Notation

0.4 < x \leq 2.66666 \dots

Solution steps

\frac{2 x ^{2} - 1}{5 x - 2} \leq \frac{4 x + 1}{10}

Rewrite in standard form

\frac{3 x - 8}{5 x - 2} \leq 0

Identify the intervals

\frac{2}{5} < x \leq \frac{8}{3}

\frac{a ^{- 16}}{a ^{- 4}}

x^{2} - 16 x + 63 > 0

\frac{x ^{4} y ^{3}}{x ^{2} y ^{3}}

- (14 x)^{0} y^{- 7} z

- 4 x = 64

What is (2x^2-1)/(5x-2)<= (4x+1)/(10) ?
The solution to (2x^2-1)/(5x-2)<= (4x+1)/(10) is 2/5 <x<= 8/3