Example Question - linear inequality
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Solving a System of Inequalities
<p>Решим систему неравенств:</p> <p>\[ \begin{align*} x^2 &< 9 \\ 2x &> 1 \end{align*} \]</p> <p>Решим первое неравенство:</p> <p>\[ x^2 < 9 \Rightarrow -3 < x < 3 \]</p> <p>Решим второе неравенство:</p> <p>\[ 2x > 1 \Rightarrow x > \frac{1}{2} \]</p> <p>Комбинируем оба неравенства:</p> <p>\[ \frac{1}{2} < x < 3 \]</p> <p>Теперь найдём наибольшее целочисленное решение системы неравенств:</p> <p>Наибольшее целое число, которое удовлетворяет условию, это \( x = 2 \).</p>
Graphical Representation of a Linear Inequality Solution
<p>Simplify the given inequality \( (3x + 1)^2 \geq (3x + 1)(3) \)</p> <p>Expand both sides: \( 9x^2 + 6x + 1 \geq 9x + 3 \)</p> <p>Rearrange terms to set the inequality to zero: \( 9x^2 + 6x + 1 - 9x - 3 \geq 0 \)</p> <p>Simplify: \( 9x^2 - 3x - 2 \geq 0 \)</p> <p>Factor the quadratic inequality: \( (3x + 1)(3x - 2) \geq 0 \)</p> <p>Identify critical points where the inequality can change signs by setting each factor equal to zero: \( 3x + 1 = 0 \) and \( 3x - 2 = 0 \)</p> <p>Solve for x to find the critical points: \( x = -\frac{1}{3} \) and \( x = \frac{2}{3} \)</p> <p>Test intervals to determine where the inequality holds; use test points \( x < -\frac{1}{3} \), \( -\frac{1}{3} < x < \frac{2}{3} \), and \( x > \frac{2}{3} \)</p> <p>Choose test points such as \( x = -1 \), \( x = 0 \), and \( x = 1 \); verify that the inequality is satisfied for \( x < -\frac{1}{3} \) and \( x > \frac{2}{3} \) but not in the middle</p> <p>The solution is \( x \in (-\infty, -\frac{1}{3}] \) union \( [\frac{2}{3}, \infty) \)</p> <p>Therefore, the correct graph is D.</p>
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