What is (n+1)/(n+2)<= (3n+2)/(2n+6) ?

The solution to (n+1)/(n+2) = sqrt(2)

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(n+1)/(n+2)<= (3n+2)/(2n+6)

\frac{n + 1}{n + 2} \leq \frac{3 n + 2}{2 n + 6}

n < - 3 or - 2 < n \leq - 2 or n \geq 2

Interval Notation

(- \infty, - 3) \cup (- 2, - 2] \cup [2, \infty)

Decimal Notation

n < - 3 or - 2 < n \leq - 1.41421 \dots or n \geq 1.41421 \dots

Solution steps

\frac{n + 1}{n + 2} \leq \frac{3 n + 2}{2 n + 6}

Rewrite in standard form

\frac{- n ^{2} + 2}{( n + 2 ) ( n + 3 )} \leq 0

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( n + 2 ) ( n - 2 ) ) ( - 1 )}{( n + 2 ) ( n + 3 )} \geq 0 \cdot (- 1)

Simplify

\frac{( n + 2 ) ( n - 2 )}{( n + 2 ) ( n + 3 )} \geq 0

Identify the intervals

n < - 3 or - 2 < n \leq - 2 or n \geq 2

y - 9 y = 0

2 x - 20 = 7 x - 5

(0.03)^{3}

∣ x + 5 ∣ \leq 0

a^{3} b^{5} c^{2} - a^{4} b^{5} c^{7} + a^{6} b c^{5}

What is (n+1)/(n+2)<= (3n+2)/(2n+6) ?
The solution to (n+1)/(n+2)<= (3n+2)/(2n+6) is n<-3\lor-2<n<=-sqrt(2)\lor n>= sqrt(2)