1.5 KiB
1. Solve for \(`x`\) if \(`\dfrac{2x+3}{4} - \dfrac{3x-1}{3} = \dfrac{5x - 2}{2}`\)
\(`3(2x+3) - 4(3x-1) = 6(5x-2)`\)
\(`6x+9 - 12x + 4 = 30x - 12`\)
\(`-6x + 13 = 30x - 12`\)
\(`36x = 25`\)
\(`x = \dfrac{25}{36}`\)
2. Determine the length of \(`AB`\) for \(`\triangle ABC`\), where \(`D`\) is on segment \(`AC`\), and \(`AD = 7, DC = 3,BC = 5`\) - \(`\because 3^2 + 4^2 = 5^2 `\) - \(`\therefore BD = 4`\) - \(`\because \angle{ADB} = 90\degree`\) - \(`\therefore BA = BD^2 + 7^2`\) - \(`\therefore BA = \sqrt{4^2 + 7^2}`\) - \(`BA = \sqrt{65}`\) - The length of \(`AB`\) is \(`\sqrt{65}`\)
3. Line 1 goes through the points \(`(-3, -7)`\) and \(`(9, 1)`\). Line 2 is perpendicular to \(`3x-4y+8 = 0`\) and has a y-intercept of \(`7`\). Determine the point of intersection of line 1 and line 2. - \(`m_1 = \dfrac{1-(-7)}{9-(-3)}`\)
\(`m_1 = \dfrac{8}{12}`\)
\(`m_1 = \dfrac{2}{3}`\)
\(`1 = \dfrac{2}{3}(9) + b`\)
\(`1 = 6 + b`\)
\(`b = -5`\)
\(`y_1 = \dfrac{2}{3}x - 5`\)
\(`3x - 4y + 8 = 0 `\)
\(`4y = 3x + 8`\)
\(`y = \dfrac{3}{4}x + 2`\)
\(`m_{\perp} = \dfrac{-4}{3}`\)
\(`y_2 = \dfrac{-4}{3}x + 7`\)
\(`\begin{cases} y = \dfrac{-4}{3}x + 7 \quad (1) \\ \\y = \dfrac{2}{3}x - 5 \quad (2) \end{cases}`\)
sub \(`(1)`\) into \(`(2)`\)
\(`\dfrac{2}{3}x - 5 = \dfrac{-4}{3}x + 7`\)
\(`\dfrac{6}{3}x = 12`\)
\(`2x = 12`\)
\(`x = 6 (3)`\)
sub \(`(3)`\) into \(`(2)`\)
\(`y = \dfrac{-24}{3} + 7`\)
\(`y = -8 + 7`\)
\(`y = -1`\)
\(`\therefore`\) the Point of Intersection is \(`(6, -1)`\)