What is (x^2+1)/(x^2-3x+2)> x/(x^2-3x+2) ?

The solution to (x^2+1)/(x^2-3x+2)> x/(x^2-3x+2) is x 2

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(x^2+1)/(x^2-3x+2)> x/(x^2-3x+2)

\frac{x ^{2} + 1}{x ^{2} - 3 x + 2} > \frac{x}{x ^{2} - 3 x + 2}

x < 1 or x > 2

Interval Notation

(- \infty, 1) \cup (2, \infty)

Solution steps

\frac{x ^{2} + 1}{x ^{2} - 3 x + 2} > \frac{x}{x ^{2} - 3 x + 2}

Rewrite in standard form

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

Factor

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

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

Identify the intervals

x < 1 or x > 2

3.6 + 7.2 n < 25.2

9 = 4 (3 k - 4) - 7 k

12 a - 16

5 x > 8

27 a^{3} + 64

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