[tex]\sqrt{-5x-4} +\sqrt{1-3x} \ \textless \ 3[/tex]

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вопрос

To solve this inequality, we need to be careful with the domain of the square roots involved. In order for the square roots to be real numbers, the expressions under the square roots must be greater than or equal to 0.

For [tex]\sqrt{-5x-4}[/tex] to be real, we must have [tex]-5x-4 \geq 0[/tex]. Solving this inequality, we get: [tex]x \leq -\frac{4}{5}[/tex].

For [tex]\sqrt{1-3x}[/tex] to be real, we must have [tex]1-3x \geq 0[/tex]. Solving this inequality, we get: [tex]x \leq \frac{1}{3}[/tex].

Now we need to consider the cases where the inequality [tex]\sqrt{-5x-4} +\sqrt{1-3x} \ \textless \ 3[/tex] holds.

Case 1: [tex]-\frac{4}{5} \leq x \leq \frac{1}{3}[/tex]
In this case, both square roots are real numbers and the inequality holds true.

Case 2: [tex]x \leq -\frac{4}{5} \ and \ x \leq \frac{1}{3}[/tex]
In this case, the expression under the square root in [tex]\sqrt{1-3x}[/tex] becomes negative, hence the inequality does not hold true.

Thus, the solution to the inequality [tex]\sqrt{-5x-4} +\sqrt{1-3x} \ \textless \ 3[/tex] is: [tex]-\frac{4}{5} \leq x \leq \frac{1}{3}[/tex].