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

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

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(y+3)/(y^2+1)>1

\frac{y + 3}{y ^{2} + 1} > 1

- 1 < y < 2

Interval Notation

(- 1, 2)

Solution steps

\frac{y + 3}{y ^{2} + 1} > 1

Rewrite in standard form

\frac{- y ^{2} + y + 2}{y ^{2} + 1} > 0

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( y + 1 ) ( y - 2 ) ) ( - 1 )}{y ^{2} + 1} < 0 \cdot (- 1)

Simplify

\frac{( y + 1 ) ( y - 2 )}{y ^{2} + 1} < 0

Identify the intervals

- 1 < y < 2

10 x^{2} + 19 x - 15

\frac{3}{5} + \frac{1}{3} > 2

\frac{3}{4} x^{2} - 4 x - 3 = 0

6 x + 1 = 2 x + 1

\frac{4 x ^{2} y z ^{5}}{6 x ^{2} y ^{3} z ^{2}}