Question

(x+y)/(ax+by)=(y+z)/(ay+bz)=(z+x)/az+bx. Prove that all equal to 2/a+b..

Answer 1

$\textbf{Concept used:}$

$\text{If}\;\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$

$\text{then}$

$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{a+c+e}{b+d+f}$

$\textbf{Given:}$

$\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}$

$\text{Then, using the given concept}$

$\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}=\frac{x+y+y+z+z+x}{ax+by+ay+bz+az+bx}$

$\implies\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}=\frac{2x+2y+2z}{a(x+y+z)+b(x+y+z)}$

$\implies\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}=\frac{2(x+y+z)}{(x+y+z)(a+b)}$

$\implies\bf\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}=\frac{2}{a+b}$

Answer 2

Answer:

Step-by-step explanation:

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