What is sqrt(2x-1)+6<= 15 ?

The solution to sqrt(2x-1)+6<= 15 is 1/2 <= x<= 41

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sqrt(2x-1)+6<= 15

2 x - 1 + 6 \leq 15

\frac{1}{2} \leq x \leq 41

Interval Notation

[\frac{1}{2}, 41]

Decimal Notation

0.5 \leq x \leq 41

Solution steps

2 x - 1 + 6 \leq 15

Move

6

to the right side

2 x - 1 \leq 9

Square both sides

(2 x - 1)^{2} \leq 9^{2}

Simplify

2 x - 1 \leq 81

2 x - 1 \leq 81 : x \leq 41

x \leq 41

Find singularity points

Find non-negative values for radicals:

x \geq \frac{1}{2}

Combine the intervals

x \leq 41 and x \geq \frac{1}{2}

Merge Overlapping Intervals

\frac{1}{2} \leq x \leq 41

\frac{x}{2} + \frac{x}{5} = \frac{7}{10}

a^{\frac{5}{4}} b^{\frac{13}{4}}

- 5 + \frac{( 2 \cdot ( 25 + 100 ) )}{3 - x} < 0

3 x^{2} - 9 x

3 s^{2} + 30 s + 63 = 0