A tricky Algebra from Oxford University Admission Interview. Entrance Aptitude Test. Find x=? & y=?

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[@pastcontrolsfuture]

4 and -3 in my head by logical deduction

[@alessiodaini7907]

2 cases: a = b, that's possible, since, at this way, the system is composed by 2 identical equations; a ≠ b. case a = b: a = b => a² = a + 13 => a² - a - 13 = 0 => a = 1/2(1±sqrt(53)) case a ≠ b a² = b + 13 & b² = a + 13 from this system is true that 13 = a² - b = b² - a, so we have: a² - b = b² - a a² - b² = b - a = -(a - b) a² - b² = (a-b)(a+b), so by substitution: (a-b)(a+b) = -(a-b) a + b = -1 => a = -(b+1) by substituting a in b² = a + 13, we obtain: b² = -b + 12 => b² + b - 12 = 0, from that b ={-4,3} so if we substitute b solutions in a = -(b+1) there's these couples of solutions: (a,b) ={(3,-4),(-4,3)} In conclusion we have these solutions, including all cases: (a,b) ={(3,-4),(-4,3),(1/2(1±sqrt(53)),1/2(1±sqrt(53)))} Yeah, I added also the other cases to find all real solutions

[@독침전략미사일주권]

I agree your solution. Because I got the result like you.

[@cabasantbab]

A= -4 B= 3

[@mrinalkulkarni3151]

Subtract both given equations a^2-b^2=b-a. i . e (a+b).(a-b)=-1(a-b) i .e a=-1-b . Now substitute this in second equation b^2=-1-b+13. i .e b^2 +b-12=0 i .e (b-4).(b+3)=0 i .e b=(4,-3) now a can be calculated

[@bobross7473]

There are two more solutions. I left a comment about it.

[@bobross7473]

Actually you’re right, I forgot he said a =/= b

[@vitalguard]

富士山 :)

[@jonathanday6692]

a^2-13=b, so (a^2-13)^2=a+13. Expanding, a^4-26a^2+169=a+13 Rearrange, giving us a^4-26a^2-a=-156 Allowing us to get a directly.

[@lalasofy]

A^2=b+13 which is a^2=16

[@passamaquoddy8311]

Your English is excellent !!!

[@superacademy247]

Thanks for your kind words

[@lalasofy]

It is tricky

[@unknownguy-x3134]

Well if this is oxford then India's 10 grade(high school final year) is way harder

[@bobross7473]

There are two irrational solutions not shown in the video. The solutions are: (a,b) = (((1+sqrt(53))/2), ((1+sqrt(53))/2)) and (((1-sqrt(53))/2), ((1-sqrt(53))/2)). You can plug these solutions back into the original equations and find that they are indeed valid solutions.

[@bobross7473]

Oh wait sorry you said they aren’t equal, my bad

[@fron3107]

When did B become P?

[@lalasofy]

So b^2 =17 because a+13=17

[@prollysine]

q=b^2-13 , (b-3)(b^3+3b^2-17b-52)=0 , b=3 , b^3+3b^2-17b-52=0 , (b+4)(b^2-b-13)=0 , b=-4 , b^2-b-13=0 , b=(1+/-V53)/2 , b= 3 , -4 , case 1 , b=3 , a=b^2-13 , a=9-13 , b= -4 , case 2 , b=-4 , a=16-13 , b=3 , solu , (a , b) , (-4 , 3) , (3 , -4) , test , a^2=b+13 , c1 , (-4)^2=3+13 , 16=16 , c2 , 3^2= -4+13 , 9=9 , OK ,

[@bobross7473]

There’s two more solutions

[@bobross7473]

Actually you are right I forgot he said a not equal to b

[@doowadiwadi]

Oxford doesn't do admissions like that. Apart from that, the question is way too easy to ever to be considered in case Oxford would do admissions like this.

[@key_board_x]

(1): a² = b + 13 (2): b² = a + 13 (1) - (2) a² - b² = (b + 13) - (a + 13) a² - b² = b + 13 - a - 13 a² - b² = b - a → recall: a² - b² = (a + b).(a - b) (a + b).(a - b) = b - a (a + b).(a - b) - (b - a) = 0 (a + b).(a - b) + (a - b) = 0 (a - b).[(a + b) + 1] = 0 → where: a ≠ b a + b + 1 = 0 a + b = - 1 ← equation (3) (1) + (2) a² + b² = (b + 13) + (a + 13) a² + b² = a + b + 26 → recall (3): a + b = - 1 a² + b² = 25 ← equation (4) From (3): a + b = - 1 (a + b)² = 1 a² + b² + 2ab = 1 → recall (4): a² + b² = 25 25 + 2ab = 1 2ab = - 24 ← equation (5) (a - b)² = a² + b² - 2ab → recall (4): a² + b² = 25 (a - b)² = 25 - 2ab → recall (5): 2ab = - 24 (a - b)² = 25 + 24 (a - b)² = 49 a - b = ± 7 ← equation (6) First case: a - b = 7 → recall (3): a + b = - 1 a + b = - 1 --------------------------the sum 2a = 6 → a = 3 → recall: a + b = - 1 → b = - 4 Second case: a - b = - 7 → recall (3): a + b = - 1 a + b = - 1 --------------------------the sum 2a = - 8 → a = - 4 → recall: a + b = - 1 → b = 3

[@bobross7473]

There are two more solutions; see my comment in here

[@bobross7473]

You are right actually, I forgot that he said a not equal to b

[@XennialGuy]

This is part of the entrance exam into Oxford? Are they letting elementary school kids apply these days? Boy is the bar set low.

[@THIPPESWAMYSAJJANGT]

Hi i have another method....

[@superacademy247]

What is it? Share with us!

[@TheNizzer]

More creative to let a=x, b =y. Sketch the two parabolas, verify two solutions (and symmetry). Not hard to realise 13+3 = 4^2 is it? Especially if as claimed, it’s for Oxford interviews. Most Oxford Maths and Physics candidates would do this in their head!

[@XennialGuy]

I solved this in my head in less than 2 minutes. A = -4 B = 3 You made this way too complicated.

[@lto1964]

You are complicating things.

[@ЭдуардПлоткин-р3л]

Неправильное решение.От начала до конца.

[@mohammadahmadi2980]

not the only solutions, you can verify

[@superacademy247]

The solutions are restricted. a is NOT equal to b