What is ((2x^2-1))/((2x^2+7x-15))>1 ?

The solution to ((2x^2-1))/((2x^2+7x-15))>1 is x<-5\lor 3/2 <x<2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((2x^2-1))/((2x^2+7x-15))>1

\frac{( 2 x ^{2} - 1 )}{( 2 x ^{2} + 7 x - 15 )} > 1

x < - 5 or \frac{3}{2} < x < 2

Interval Notation

(- \infty, - 5) \cup (\frac{3}{2}, 2)

Decimal Notation

x < - 5 or 1.5 < x < 2

Solution steps

\frac{2 x ^{2} - 1}{2 x ^{2} + 7 x - 15} > 1

Rewrite in standard form

\frac{- 7 x + 14}{2 x ^{2} + 7 x - 15} > 0

Factor

\frac{- 7 x + 14}{2 x ^{2} + 7 x - 15} : \frac{- 7 ( x - 2 )}{( 2 x - 3 ) ( x + 5 )}

\frac{- 7 ( x - 2 )}{( 2 x - 3 ) ( x + 5 )} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 7 ( x - 2 ) ) ( - 1 )}{( 2 x - 3 ) ( x + 5 )} < 0 \cdot (- 1)

Simplify

\frac{7 ( x - 2 )}{( 2 x - 3 ) ( x + 5 )} < 0

Divide both sides by

7

\frac{\frac{7 ( x - 2 )}{( 2 x - 3 ) ( x + 5 )}}{7} < \frac{0}{7}

Simplify

\frac{x - 2}{( 2 x - 3 ) ( x + 5 )} < 0

Identify the intervals

x < - 5 or \frac{3}{2} < x < 2

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

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

(a + 3) (a - 2) < 0

\frac{2}{( x - 2 )} \geq \frac{1}{x}

x^{2} \geq - 5

What is ((2x^2-1))/((2x^2+7x-15))>1 ?
The solution to ((2x^2-1))/((2x^2+7x-15))>1 is x<-5\lor 3/2 <x<2