What is (2x-1)/3 >x-5\land 3x+4<(7x+9)/2 ?

The solution to (2x-1)/3 >x-5\land 3x+4<(7x+9)/2 is -1<x<14

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(2x-1)/3 >x-5\land 3x+4<(7x+9)/2

\frac{2 x - 1}{3} > x - 5 and 3 x + 4 < \frac{7 x + 9}{2}

- 1 < x < 14

Interval Notation

(- 1, 14)

Solution steps

\frac{2 x - 1}{3} > x - 5 and 3 x + 4 < \frac{7 x + 9}{2}

\frac{2 x - 1}{3} > x - 5 : x < 14

3 x + 4 < \frac{7 x + 9}{2} : x > - 1

Combine the intervals

x < 14 and x > - 1

Merge Overlapping Intervals

- 1 < x < 14

1 - (\frac{17}{20})^{2}

5 y + 3 > - 7 y + 13

n^{2} + 3 n - 12 = 6

∣ 9 x - 8 ∣ = 0

16 z^{2} + 61 = 64 z

What is (2x-1)/3 >x-5\land 3x+4<(7x+9)/2 ?
The solution to (2x-1)/3 >x-5\land 3x+4<(7x+9)/2 is -1<x<14