Help me!

Can someone help me to solve this problem? $a, b, c>0$

$\frac{3b+5c}{a} = \frac{3a+4c}{b} = \frac{5a+4b}{c}$

Find $\large a:b:c$ . (Or actually the answer doesn't exist?)

#Algebra

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`2 \times 3`	$2 \times 3$
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Comments

Hint: Set $a = 1$ , you can now obtain the values of $b$ and $c$ .

Hint 2: Set $a=2$ , you can now obtain the values of $b$ and $c$ again.

Hint 3: Why is it the case that when $a=1$ or when $a=2$ or for any other positive value of $a$ , the ratio $a:b:c$ remains the same?

Hint 4: Let $a + b + c = k$ for some positive $k$ .

Pi Han Goh - 3 years, 6 months ago

Thanks! I get it!

(But it is hard to find $b, c$ when $a=1$ .)

Chan Tin Ping - 3 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$