Question

$\huge \sf \color{red}{\underbrace{Solve}} :$

If a:b = c:d, show that $\large \frac{2a + 3b}{2c+3d} = \frac{2a-3b}{2c-3d}$​

Answer 1

$\huge \sf \color{blue}{\underline{\underline{Solution}}} :$

$\large \sf \frac{a}{b}=\frac{c}{d}$

$\colorbox{skyblue}{ \sf Multiplying both by ⅔ }$

$\large \sf \implies \frac{2a}{3b}= \frac{2c}{3d}$ $\\ \\ \colorbox{skyblue}{By \ componendo \ and \ dividendo } \\ \\ \implies \frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d} \\ \\ \implies \frac{2a+3b}{2c+3d} = \frac{2a-3b}{2c+3d}$

$\huge \underline{Hence \: \: proved}$

Answer 2

Answer:

$\huge \sf \color{blue}{\underline{\underline{Solution}}} :Solution:</p><p></p><p>\large \sf \frac{a}{b}=\frac{c}{d}ba=dc</p><p></p><p>\colorbox{skyblue}{ \sf Multiplying both by ⅔ } Multiplying both by ⅔ </p><p></p><p>\large \sf \implies \frac{2a}{3b}= \frac{2c}{3d}⟹3b2a=3d2c \begin{gathered} \\ \\ \colorbox{skyblue}{By \ componendo \ and \ dividendo } \\ \\ \implies \frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d} \\ \\ \implies \frac{2a+3b}{2c+3d} = \frac{2a-3b}{2c+3d} \end{gathered}By  componendo  and  dividendo ⟹2a−3b2a+3b=2c−3d2c+3d⟹2c+3d2a+3b=2c+3d2a−3b</p><p></p><p>\huge \underline{Hence \: \: proved}Henceproved</p><p></p><p>$