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Bunuel
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Hello KanishkM !

Can someone please give me some advice on how to improve the "way" to approach this kind of exercises?

In the end, I made it correct but I was stuck several minutes doing it in the wrong way, I was expanding the term:

(x + y)^2 = 9 and x + 2y = 2

x^2 + 2xy + y^2 = 9

x2 + y (2x + y) = 9

At this point, I thought I was going well but I couldn't get further and then the time went against me and I got stressed and stopped thinking.

Thank you so much in advance!

Thank you KanishkM narendran1990 !
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Hello Jfrancis,

I solved the question in the same way you have done, though it doesn't yield a quick solution, it's solvable:

Taking from your last step - x^2 + y^2 + 2y.x = 9

x^2 + y^2+(2-x).x=9 [x+2y=2, rewritten as 2y=2-x)

Upon expansion, x^2 cancels out giving the equation, 2x+y^2=9

2(2-2y)+y^2=9

4-4y+y^2 = 9

y^2-4y = 5 , taking y as common , y(y-4)=5, hence y can be 5 or 9. Only 5 is in the answer choice.
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jfranciscocuencag
Hello KanishkM !

Can someone please give me some advice on how to improve the "way" to approach this kind of exercises?

In the end, I made it correct but I was stuck several minutes doing it in the wrong way, I was expanding the term:

(x + y)^2 = 9 and x + 2y = 2

x^2 + 2xy + y^2 = 9

x2 + y (2x + y) = 9

At this point, I thought I was going well but I couldn't get further and then the time went against me and I got stressed and stopped thinking.

Thank you so much in advance!

jfranciscocuencag

So i believe you got the response on how this question could have been solved by using your approach.

Whenever you get such equations with two vairables, try to reduce the question in simplest form. If after substitution you are able to get an easy expression, you are on right track, if not, then you missed out on something.

Eg : just for illustration
2x -3y = 6, find x

y = (2x-6) /3

Whenever you think you are taking time to solve the question, you assumed something, in turn made the question complicated.

Posted from my mobile device
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