What is 7sqrt(x)+1<= 9 ?

The solution to 7sqrt(x)+1<= 9 is 0<= x<= 64/49

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7sqrt(x)+1<= 9

7 x + 1 \leq 9

0 \leq x \leq \frac{64}{49}

Interval Notation

[0, \frac{64}{49}]

Decimal Notation

0 \leq x \leq 1.30612 \dots

Solution steps

7 x + 1 \leq 9

Move

1

to the right side

7 x \leq 8

Divide both sides by

7

x \leq \frac{8}{7}

Square both sides

(x)^{2} \leq (\frac{8}{7})^{2}

Simplify

x \leq (\frac{8}{7})^{2}

Expand

(\frac{8}{7})^{2} : \frac{64}{49}

x \leq \frac{64}{49}

Find singularity points

Find non-negative values for radicals:

x \geq 0

Combine the intervals

x \leq \frac{64}{49} and x \geq 0

Merge Overlapping Intervals

0 \leq x \leq \frac{64}{49}

\frac{2 x y ^{- 3}}{x ^{- 2} y ^{4}}

\frac{6.5522 + 2.23}{0.1234}

81 x^{2} - 16

4 x = - 5

∣ x - 3 ∣ - 5 = - 4