Question

Divide 28 into two parts in such a way that 6/5 of one part is equal to 2/3 of the other?

Answer 1

ANSWER WITH FULL EXPLANATION. 
Let one part be $x$. 
The other part will be $(28-x)$. 
Now, 
ATQ (According To Question)
6/5 of one part = 2/3 of the other part
⇒ $\frac{6}{5}x = \frac{2}{3}(28-x)$
⇒ $\frac{6}{5}x = \frac{2(28-x)}{3}$
⇒ $\frac{6}{5}x = \frac{56-2x}{3}$
Multiplying both sides of the equation by 15, the LCM of the denominators 5 and 3. 
⇒ $\frac{6}{5}x*15 = \frac{56-2x}{3}*15$
⇒ $18x = 5(56-2x)$
⇒ $18x = 280-10x$
Adding 10x to both sides. 
⇒ $18x+10x = 280-10x+10x$
⇒ $28x = 280$
Dividing both sides of the equation by 28. 
⇒ $\frac{28x}{28} = \frac{280}{28}$
⇒ $x = 10$

x = 10
(28-x) = (28-10) = 18
∴ 28 can be divided into two parts, 10 and 18, according to the criteria (condition) given in your question. 

Congrats! Solution found. 

Hope that this may help you. 

If you have any doubt, then you can ask me in the comments. 

Answer 2

Answer:

Step-by-step explanation:

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