Question

if 4x-3z/4c=4-3y/3b=4y-3x/2a show that each ratio =x+y+z/2a+3b+4c​

Answer 1

$\textbf{Given:}$

$\dfrac{4x-3z}{4c}=\dfrac{4z-3y}{3b}=\dfrac{4y-3x}{2a}$

$\textbf{To prove:}$

$\text{Each ratio}=\dfrac{x+y+z}{2a+3b+4c}$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{4x-3z}{4c}=\dfrac{4z-3y}{3b}=\dfrac{4y-3x}{2a}$

$\text{We know that,}$

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

$\text{then,}\;\bf\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{a+c+e}{b+d+f}$

$\implies\dfrac{4x-3z}{4c}=\dfrac{4z-3y}{3b}=\dfrac{4y-3x}{2a}=\dfrac{4x-3z+4z-3y+4y-3x}{4c+3b+2a}$

$\implies\dfrac{4x-3z}{4c}=\dfrac{4z-3y}{3b}=\dfrac{4y-3x}{2a}=\dfrac{x+y+z}{4c+3b+2a}$

$\implies\bf\dfrac{4x-3z}{4c}=\dfrac{4z-3y}{3b}=\dfrac{4y-3x}{2a}=\dfrac{x+y+z}{2a+3b+4c}$

Find more:

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

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Answer 2

Answer:

Step-by-step explanation: