What is (6x^2-17x+12)/(4x^2+2x-12)<= 0 ?

The solution to (6x^2-17x+12)/(4x^2+2x-12)<= 0 is -2<x<= 4/3

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(6x^2-17x+12)/(4x^2+2x-12)<= 0

\frac{6 x ^{2} - 17 x + 12}{4 x ^{2} + 2 x - 12} \leq 0

- 2 < x \leq \frac{4}{3}

Interval Notation

(- 2, \frac{4}{3}]

Decimal Notation

- 2 < x \leq 1.33333 \dots

Solution steps

\frac{6 x ^{2} - 17 x + 12}{4 x ^{2} + 2 x - 12} \leq 0

Factor

\frac{6 x ^{2} - 17 x + 12}{4 x ^{2} + 2 x - 12} : \frac{( 3 x - 4 ) ( 2 x - 3 )}{2 ( 2 x - 3 ) ( x + 2 )}

\frac{( 3 x - 4 ) ( 2 x - 3 )}{2 ( 2 x - 3 ) ( x + 2 )} \leq 0

Multiply both sides by

2

\frac{2 ( 3 x - 4 ) ( 2 x - 3 )}{2 ( 2 x - 3 ) ( x + 2 )} \leq 0 \cdot 2

Simplify

\frac{3 x - 4}{x + 2} \leq 0

Identify the intervals

- 2 < x \leq \frac{4}{3}

j^{2} - 10 j + 9

\frac{1}{i ^{2}}

- 2 x^{2} - 5 x - 4 = 0

x^{2} = 44

x^{2} + 5^{2} = 6^{2}

What is (6x^2-17x+12)/(4x^2+2x-12)<= 0 ?
The solution to (6x^2-17x+12)/(4x^2+2x-12)<= 0 is -2<x<= 4/3