0000:00:00:00:00Start25Number 25ExplanationThe solution set of ∣x3−14∣=112\left|\frac{x}{3} - \frac{1}{4}\right| = \frac{1}{12}3x−41=121 is....{−12,2}\{-\frac{1}{2}, 2\}{−21,2}{2,3}\{2, 3\}{2,3}{1,2}\{1, 2\}{1,2}{−32,13}\{-\frac{3}{2}, \frac{1}{3}\}{−23,31}{12,1}\{\frac{1}{2}, 1\}{21,1}ExplanationFirst change it so that the fractional form disappears ∣x3−14∣=112\left|\frac{x}{3} - \frac{1}{4}\right| = \frac{1}{12}3x−41=121∣4x−312∣=112\left|\frac{4x - 3}{12}\right| = \frac{1}{12}124x−3=121∣4x−3∣=1|4x - 3| = 1∣4x−3∣=1 Its definition is ∣4x−3∣={4x−3,for x≥34−(4x−3),for x<34|4x - 3| = \begin{cases} 4x - 3, & \text{for } x \geq \frac{3}{4} \\ -(4x - 3), & \text{for } x < \frac{3}{4} \end{cases}∣4x−3∣={4x−3,−(4x−3),for x≥43for x<43 For x≥34x \geq \frac{3}{4}x≥43 then ∣4x−3∣=1|4x - 3| = 1∣4x−3∣=14x−3=14x - 3 = 14x−3=14x=44x = 44x=4x=1x = 1x=1 For x<34x < \frac{3}{4}x<43 then ∣4x−3∣=1|4x - 3| = 1∣4x−3∣=1−(4x−3)=1-(4x - 3) = 1−(4x−3)=1−4x+3=1-4x + 3 = 1−4x+3=1−4x=−2-4x = -2−4x=−2x=12x = \frac{1}{2}x=21 So the solution set is {12,1}\left\{\frac{1}{2}, 1\right\}{21,1}.
25Number 25ExplanationThe solution set of ∣x3−14∣=112\left|\frac{x}{3} - \frac{1}{4}\right| = \frac{1}{12}3x−41=121 is....{−12,2}\{-\frac{1}{2}, 2\}{−21,2}{2,3}\{2, 3\}{2,3}{1,2}\{1, 2\}{1,2}{−32,13}\{-\frac{3}{2}, \frac{1}{3}\}{−23,31}{12,1}\{\frac{1}{2}, 1\}{21,1}ExplanationFirst change it so that the fractional form disappears ∣x3−14∣=112\left|\frac{x}{3} - \frac{1}{4}\right| = \frac{1}{12}3x−41=121∣4x−312∣=112\left|\frac{4x - 3}{12}\right| = \frac{1}{12}124x−3=121∣4x−3∣=1|4x - 3| = 1∣4x−3∣=1 Its definition is ∣4x−3∣={4x−3,for x≥34−(4x−3),for x<34|4x - 3| = \begin{cases} 4x - 3, & \text{for } x \geq \frac{3}{4} \\ -(4x - 3), & \text{for } x < \frac{3}{4} \end{cases}∣4x−3∣={4x−3,−(4x−3),for x≥43for x<43 For x≥34x \geq \frac{3}{4}x≥43 then ∣4x−3∣=1|4x - 3| = 1∣4x−3∣=14x−3=14x - 3 = 14x−3=14x=44x = 44x=4x=1x = 1x=1 For x<34x < \frac{3}{4}x<43 then ∣4x−3∣=1|4x - 3| = 1∣4x−3∣=1−(4x−3)=1-(4x - 3) = 1−(4x−3)=1−4x+3=1-4x + 3 = 1−4x+3=1−4x=−2-4x = -2−4x=−2x=12x = \frac{1}{2}x=21 So the solution set is {12,1}\left\{\frac{1}{2}, 1\right\}{21,1}.
