d. (7-2 x)/(2) (2)/(5)(2-x)=1 (1)/(4)

To solve the equation \(\frac{7-2x}{2} \cdot \frac{2}{5}(2-x) = 1\frac{1}{4}\), follow these steps:<br /><br />1. Convert the mixed number to an improper fraction:<br /> \[<br /> 1\frac{1}{4} = \frac{5}{4}<br /> \]<br /><br />2. Rewrite the equation with the improper fraction:<br /> \[<br /> \frac{7-2x}{2} \cdot \frac{2}{5}(2-x) = \frac{5}{4}<br /> \]<br /><br />3. Simplify the left-hand side:<br /> \[<br /> \frac{7-2x}{2} \cdot \frac{2(2-x)}{5} = \frac{5}{4}<br /> \]<br /> \[<br /> \frac{(7-2x)(2-x)}{5} = \frac{5}{4}<br /> \]<br /><br />4. Multiply both sides by 5 to clear the denominator:<br /> \[<br /> (7-2x)(2-x) = \frac{5}{4} \cdot 5<br /> \]<br /> \[<br /> (7-2x)(2-x) = \frac{25}{4}<br /> \]<br /><br />5. Expand the left-hand side:<br /> \[<br /> 14 - 7x - 4x + 2x^2 = \frac{25}{4}<br /> \[<br />2x^2 - 11x + 14 = \frac{25}{4}<br /> \]<br /><br />6. Multiply both sides by 4 to clear the fraction:<br /> \[<br /> 4(2x^2 - 11x + 14) = 25<br /> \]<br /> \[<br /> 8x^2 - 44x + 56 = 25<br /> \]<br /><br />7. Subtract 25 from both sides to set the equation to zero:<br /> \[<br /> 8x^2 - 44x + 31 = 0<br /> \]<br /><br />8. Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 8\), \(b = -44\), and \(c = 31\):<br /> \[<br /> x = \frac{-(-44) \pm \sqrt{(-44)^2 - 4 \cdot 8 \cdot 31}}{2 \cdot 8}<br /> \]<br /> \[<br /> x = \frac{44 \pm \sqrt{1936 - 992}}{16}<br /> \]<br /> \[<br /> x = \frac{44 \pm \sqrt{944}}{16}<br /> \]<br /> \[<br /> x = \frac{44 \pm 2\sqrt{236}}{16}<br /> \]<br /> \[<br /> x = \frac{22 \pm \sqrt{236}}{8}<br /> \]<br /><br />9. Simplify further if possible:<br /> \[<br /> x = \frac{22 \pm \sqrt{4 \cdot 59}}{8}<br /> \]<br /> \[<br /> x = \frac{22 \pm 2\sqrt{59}}{8}<br /> \]<br /> \[<br /> x = \frac{11 \pm \sqrt{59}}{4}<br /> \]<br /><br />Thus, the solutions are:<br />\[<br />x_1 = \frac{11 - \sqrt{59}}{4}, \quad x_2 = \frac{11 + \sqrt{59}}{4}<br />\]