What is (2x-1)/(x+1)+(3x-1)/(x+2)<4+(x-7)/(x-1) ?

The solution to (2x-1)/(x+1)+(3x-1)/(x+2) 5

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(2x-1)/(x+1)+(3x-1)/(x+2)<4+(x-7)/(x-1)

\frac{2 x - 1}{x + 1} + \frac{3 x - 1}{x + 2} < 4 + \frac{x - 7}{x - 1}

- 2 < x < - \frac{5}{4} or - 1 < x < 1 or x > 5

Interval Notation

(- 2, - \frac{5}{4}) \cup (- 1, 1) \cup (5, \infty)

Decimal Notation

- 2 < x < - 1.25 or - 1 < x < 1 or x > 5

Solution steps

\frac{2 x - 1}{x + 1} + \frac{3 x - 1}{x + 2} < 4 + \frac{x - 7}{x - 1}

Rewrite in standard form

\frac{- 4 x ^{2} + 15 x + 25}{( x + 1 ) ( x + 2 ) ( x - 1 )} < 0

Factor

\frac{- 4 x ^{2} + 15 x + 25}{( x + 1 ) ( x + 2 ) ( x - 1 )} : \frac{- ( 4 x + 5 ) ( x - 5 )}{( x + 1 ) ( x + 2 ) ( x - 1 )}

\frac{- ( 4 x + 5 ) ( x - 5 )}{( x + 1 ) ( x + 2 ) ( x - 1 )} < 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( 4 x + 5 ) ( x - 5 ) ) ( - 1 )}{( x + 1 ) ( x + 2 ) ( x - 1 )} > 0 \cdot (- 1)

Simplify

\frac{( 4 x + 5 ) ( x - 5 )}{( x + 1 ) ( x + 2 ) ( x - 1 )} > 0

Identify the intervals

- 2 < x < - \frac{5}{4} or - 1 < x < 1 or x > 5

12 a^{2} - 15 ab - 30 b + 24 a

\frac{x - 3}{x + 5} \leq 0

2 x^{2} - 15 x + 18

5 x - 4 = 2 x + 5

\frac{\frac{11.97}{0.0125 ( 29 \cdot 1 0 ^{6} )}}{4}

What is (2x-1)/(x+1)+(3x-1)/(x+2)<4+(x-7)/(x-1) ?
The solution to (2x-1)/(x+1)+(3x-1)/(x+2)<4+(x-7)/(x-1) is -2<x<-5/4 \lor-1<x<1\lor x>5