What is ((2x+5))/2 <= 3+x ?

The solution to ((2x+5))/2 <= 3+x is True for all x

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((2x+5))/2 <= 3+x

\frac{( 2 x + 5 )}{2} \leq 3 + x

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

\frac{( 2 x + 5 )}{2} \leq 3 + x

Multiply both sides by

2

\frac{2 x + 5}{2} \cdot 2 \leq 3 \cdot 2 + x \cdot 2

Simplify

2 x + 5 \leq 6 + 2 x

Move

5

to the right side

2 x \leq 2 x + 1

Move

2 x

to the left side

0 \leq 1

Therefore, the solution is

True for all x

1 1^{2} + 6^{2} = x^{2}

x^{2} - 9 x + 18 \geq 0

(- 1 b - 2)^{3}

\frac{( 2 - i )}{( 2 + i )}

\frac{40 i}{3 + i}