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x\in (-\infty,-1)\cup [-\frac{19}{24},\infty)
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Algebra
\frac{5}{3x+3} \leq 8
Podobne težave pri spletnem iskanju
How do you write the solution in interval notation, and graph \displaystyle-\frac{{5}}{{3}}{x}\le-{10} ?
https://socratic.org/questions/how-do-you-write-the-solution-in-interval-notation-and-graph-5-3x-10
Interval notation: \displaystyle{\left[{6},∞\right)} See the graph below. Explanation: First, solve for x: \displaystyle-\frac{{5}}{{3}}{x}\le-{10} When dividing or multiplying by a ...
How do you solve \displaystyle\frac{{{x}-{4}}}{{{x}+{3}}}\le{0} using a sign chart?
https://socratic.org/questions/how-do-you-solve-x-4-x-3-0-using-a-sign-chart
The answer is \displaystyle{x}\in{]}-{3},{4}{]} Explanation: As you canot divide by \displaystyle{0} , therefore \displaystyle{x}\ne-{3} Let \displaystyle{f{{\left({x}\right)}}}=\frac{{{x}-{4}}}{{{x}+{3}}} ...
How do you solve \displaystyle{3}\leq{\frac{{-{6}+{x}}}{{{2}}}} ?
https://socratic.org/questions/how-do-you-solve-3-leq-frac-6-x-2
\displaystyle{12}\le{x} Explanation: Multiply both sides by 2: \displaystyle{6}\le-{6}+{x} ...add 6 to both sides: \displaystyle{12}\le{x} and BAM. You can verify this in your mind ...
How do you solve \displaystyle{3}\leq{\frac{{{2}-{x}}}{{{x}+{2}}}} ?
https://socratic.org/questions/how-do-you-solve-3-leq-frac-2-x-x-2
The solution is \displaystyle{x}\in{\left(-{2},-{1}\right]} Explanation: We rewrite the inequality as \displaystyle{3}\le\frac{{{2}-{x}}}{{{x}+{2}}} \displaystyle{3}-\frac{{{2}-{x}}}{{{x}+{2}}}\le{0} ...
How do you solve \displaystyle{\frac{{{1}}}{{{2}}}}{x}+{5}\leq{8} ?
https://socratic.org/questions/how-do-you-solve-frac-1-2-x-5-leq-8
\displaystyle\frac{{1}}{{2}}{x}+{5}\le{8} \displaystyle\Rightarrow{x}\le{6} Explanation: The algebraic rules for solving inequalities are almost exactly the same as for solving equations. ...
How do you solve \displaystyle\frac{{{x}+{3}}}{{x}}\le-{2} using a sign chart?
https://socratic.org/questions/how-do-you-solve-x-3-x-2-using-a-sign-chart
The answer is \displaystyle{x}\in{\left[-{1},{0}{[}\right.} Explanation: We cannot cross over. Let's rewrite the inequality \displaystyle\frac{{{x}+{3}}}{{x}}+{2}\le{0} \displaystyle\frac{{{x}+{3}+{2}{x}}}{{{x}}}\le{0} ...
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Podobne težave
3x+4>6
x+y<0
5 > 2x + 3
-2 < 3x+2 < 8
2x^2 \geq 50
\frac{5}{3x+3} \leq 8
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