What is (4(x^2-1))/(x-3)<= x+1 ?

The solution to (4(x^2-1))/(x-3)<= x+1 is x<=-1\lor 1/3 <= x<3

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(4(x^2-1))/(x-3)<= x+1

\frac{4 ( x ^{2} - 1 )}{x - 3} \leq x + 1

x \leq - 1 or \frac{1}{3} \leq x < 3

Solution steps

\frac{4 ( x ^{2} - 1 )}{x - 3} \leq x + 1

Rewrite in standard form

\frac{3 x ^{2} + 2 x - 1}{x - 3} \leq 0

Factor

\frac{3 x ^{2} + 2 x - 1}{x - 3} : \frac{( 3 x - 1 ) ( x + 1 )}{x - 3}

\frac{( 3 x - 1 ) ( x + 1 )}{x - 3} \leq 0

Identify the intervals

x \leq - 1 or \frac{1}{3} \leq x < 3

\frac{3 20}{2 4}

\frac{- 4}{5}

(\frac{15}{17})^{2} - (\frac{8}{17})^{2}

9 x + 6 y

5 (2 y - 9) \geq 6 (3 y + 4)

What is (4(x^2-1))/(x-3)<= x+1 ?
The solution to (4(x^2-1))/(x-3)<= x+1 is x<=-1\lor 1/3 <= x<3