What is y+10>= 13\lor (5y+1)/2 <3 ?

The solution to y+10>= 13\lor (5y+1)/2 = 3

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y+10>= 13\lor (5y+1)/2 <3

y + 10 \geq 13 or \frac{5 y + 1}{2} < 3

y < 1 or y \geq 3

Interval Notation

(- \infty, 1) \cup [3, \infty)

Solution steps

y + 10 \geq 13 or \frac{5 y + 1}{2} < 3

y + 10 \geq 13 : y \geq 3

\frac{5 y + 1}{2} < 3 : y < 1

Combine the intervals

y \geq 3 or y < 1

Merge Overlapping Intervals

y < 1 or y \geq 3

3 x (x + 1) + 7 (x + 1)

\frac{2 b ^{\frac{1}{3}}}{3 b ^{\frac{4}{3}}}

9 x^{2} + 42 x + 24

- 1 - 10 a < - 1

(2^{2})^{3}

What is y+10>= 13\lor (5y+1)/2 <3 ?
The solution to y+10>= 13\lor (5y+1)/2 <3 is y<1\lor y>= 3