Maths cse

Question

How to solve these type of equations?
Please provide step by step solution.

$2x^2+3y^2 = 275$

$2x+3y=35$

1. $x >4$
2. $x=4$
3. $x<4$
4. $x>y, x<y, x=y$

Comments

in the given equation

$2x + 3y = 35$

here x is the multiple of 2 and y is the multiple of 3, break 35 in such a way so that it satisfy the equation

$2x + 3y = 20 + 15$

$2x + 3y = 2*10 + 3*5$

on comparing we get, $x = 10$ and $y = 5$

on putting the value of x and y in $2x^2 + 3x^2 = 275$, we get

$2*{10}^2 + 3*{5}^2 = 2*100 + 3*25 = 200 + 275$

Hence $x = 10, y = 5$

$ x > y $

One more case if $x = 4$ and $y = 9$

this will also satisfy the equation

in this case $x<y$

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This is not correct method or answer .

I am also confuse that what is the correct answer and how to solve these type of questions?

I tried  the same method but when I put the value x=4 and y=9 the result is opposite and given answer is x<y .

By puting  above value I got x<y .

And by putting x=10 and y=5 I got x>y .

Then how could be the answer x<y ?

Actually I want to know the process that I solve these type question easily and quickly and most important accurately.

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Yes,this is not the correct sol,  skipped the other cases

Answer 1

option d

Comments

I think hit and trial method is the best, just need to check 3 cases

$1. x > y$

$2.x < y$

$3. x = y$

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 Harikesh Kumar  using shortcut you  may get wrong solution.

if your calculation is fast then this method is also take maximum 3-4 min.

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You are right. But accerecy is more important in gate