Solucionar 6-x/x-2leq4-x/x+2 | Microsoft Math Solver

Resolver para x

x \in (- 2, 2) \cup [10, \infty)

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5 problemas similares a:

\frac{6 - x}{x - 2} \leq \frac{4 - x}{x + 2}

Problemas similares de búsqueda web

How do you solve

- \frac{x}{6} + \frac{2 2}{6} \leq - \frac{6 x}{3}

https://socratic.org/questions/how-do-you-solve-frac-x-6-frac-22-6-leq-frac-6x-3

x \leq - 2

multiply ALL terms on both sides of the inequality by the

lowest common multiple of 6 and 3

the lowest common multiple of 6 and 3 is 6

Solving the inequality

\frac{x}{x + 1} - \frac{1}{x - 3} - 2 \leq 0

https://math.stackexchange.com/questions/2366556/solving-the-inequality-fracxx1-frac1x-3-2-leq-0

How do you solve

\frac{x}{x - 3} \leq - \frac{8}{x - 6}

using a sign chart?

https://socratic.org/questions/how-do-you-solve-x-x-3-8-x-6-using-a-sign-chart

The solution is

x \in [- 6, 3 [\cup [4, 6 [

Explanation: We cannot do crossing over

\frac{x}{x - 3} \leq - \frac{8}{x - 6}

How do you solve

\frac{2 x - 1}{x + 2} - \frac{1}{x} \leq 1

https://socratic.org/questions/how-do-you-solve-frac-2x-1-x-2-frac-1-x-leq-1

x : (- 2, 2 - 6] \cup (0, 2 + 6]

\frac{2 x - 1}{x + 2} - \frac{1}{x} \leq 1

\frac{x}{x} \cdot \frac{2 x - 1}{x + 2} - \frac{x + 2}{x + 2} \cdot \frac{1}{x} \leq 1

How do you solve

- \frac{3}{x + 7} \leq - \frac{4}{x + 8}

https://socratic.org/questions/how-do-you-solve-3-x-7-4-x-8

The solution is

x \in (- \infty, - 8) \cup (- 7, - 4]

Explanation: We cannot do crossing over

- \frac{3}{x + 7} \leq - \frac{4}{x + 8}

How do you solve

- \frac{7}{x + 5} \leq - \frac{8}{x + 6}

https://socratic.org/questions/how-do-you-solve-7-x-5-8-x-6

{x < - 6} \cup {- 5 < x \leq 2}

- \frac{7}{x + 5} \leq - \frac{8}{x + 6}

Let's multiply ...

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