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

The solution to 1/(x-7)+1/(x+2)>= (2x)/(x^2+x-2) is x 7

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

\frac{1}{x - 7} + \frac{1}{x + 2} \geq \frac{2 x}{x ^{2} + x - 2}

x < - 2 or - \frac{5}{7} \leq x < 1 or x > 7

Interval Notation

(- \infty, - 2) \cup [- \frac{5}{7}, 1) \cup (7, \infty)

Decimal Notation

x < - 2 or - 0.71428 \dots \leq x < 1 or x > 7

Solution steps

\frac{1}{x - 7} + \frac{1}{x + 2} \geq \frac{2 x}{x ^{2} + x - 2}

Rewrite in standard form

\frac{7 x + 5}{( x - 7 ) ( x + 2 ) ( x - 1 )} \geq 0

Identify the intervals

x < - 2 or - \frac{5}{7} \leq x < 1 or x > 7

4^{1}

T + 3 \frac{T}{2} + \frac{( T + 2 )}{2} T = 4 T^{2}

2 x - 3 (x + 1) \geq 3 x

(n^{4})^{\frac{3}{2}}

- 4 - 4 (2 p + 3) = - 3 (4 p - 4)

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