Math Olympiad: How to Solve a System of Simultaneous Equations | Algebra Identities: Tricks to Know

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[rcnayak_]

Appears to be a lengthy method. Here is one easy method. Let x^2 = a and y^2 = b. So that a - b = 24. Again ab = x^2 .y^2 = (xy)^2 = 35^2 = 1225. We now from algebraic formula (a + b)^2 = (a - b)^2 + 4ab. So that (a + b)^2 = 24^2 + 4.1225 = 5476 = 74^2. Now putting the values of a and b, we get (x^2 + y^2)^2 = 74^2. So x^2 + y^2 = 74 (we can not take -74 as the left side is sum of two squares term). Now x^2 + y^2 + 2xy = 74 + 2. 35 = 144. That is (x + y)^2 = 144. So our x + y = (+/ - ) 12

[miguelgnievesl]

We have X² -Y² = 24 (Eq. 1) and XY = 35 (Eq. 2). So, Y=35/X, we substitute in Eq. 1. Then X² - (35/X)² = 24. Solving we get X⁴ -24X² - 1225 = 0. We factor (X² -49)(X² +25)=0. Then X=±7 and Y=±5. So X + Y = ±12.

[quintaldecasa]

Just by looking at the numbers, you can deduce 7 x 5 = 35. And by proving 7**2 - 5**2, you confirm. But it's good to see a complete answer by a didactic method to solve it.

[az]

Translated from Japanese
In Japan, integer problems are often encountered in high school.
xy=35
±35・±1
±7・±5
(the opposite is also possible)
x-y=±34, ±2
Since x+y is an integer
x+y=24/(x-y)
= 24/±2=±12 (answer)

[jim]

By inspection x = 7 and y = 5 (and their negatives since we're squaring). So x + y = 12 or -12. First clue: there's only so many ways to get to 35 with xy (assuming we're dealing with integers).

[kevinstreeter]

12. If this is a test, you can make some assumptions to answer quickly. Granted, you cannot do this in real life. My assumptions are that x and y are factors of 35 and x > y. Try x =7 and y = 5. 7^2 - 5^2 = 49 -25 = 24. x + y = 7 + 5 =12. It took me less than 5 seconds, which helps in a test where time is limited. Edit: x= -7 and y = -5 are also solutions.

[thesfsplayer]

35 is only can be found at 7x5
So the answer is tested after 49-25=24
The X + Y = 12

[BRUBRUETNONO]

An elegant way to solve this system is to use complex numbers to avoid solving quadratic equation twice. Here we go. Considering x and y as real numbers.
Set z=x+iy so z^2=x^2-y^2+2ixy=24+70i
Then |z^2|=√(24^2+70^2)=74

Using z_=x-iy conjugate of z,
We have z+iz_=x+iy+ix-i^2y
So z+iz_=(x+y)(1+i)
Then |z+iz_|^2=2|x+y|^2
Moreover, |z+iz_|^2=|(z+iz_)^2|
So 2|x+y|^2=|z^2-z_^2+2izz_|
=|z^2-z^2_+2i|z|^2|
=|2i.Im(z^2)+2i|z|^2|
=|2i.(Im(z^2)+|z|^2|
=|2i|.|Im(z^2)+|z|^2|
=2.|Im(z^2)+|z|^2|
As Im(Z)= imaginary part and. |Z| module of complex Z.
Then
|x+y|^2=|Im(z^2)+|z^2||
So
x+y=+/-√|Im(z^2)+|z^2||
Knowing that z^2=24+70i and |z^2|=74 then
x+y=+/-√(70+74)=+/-√144
So
x+y=+/-12
Greetings

[sawyerw]

multiply equation 1 by x^2 and substitute (xy)^2-> x^4-24x^2-35^2=0; solve quadratically for x^2= (24+/- sqrt(24^2+70^2))/2; x^2=12 +/- 37=49 or -25, x= +/-7 or +/- 5i; applying y=35/x you get all the results for x+y

[MrLidless]

All you had to do was:
(x² - y²)² + 4x²y² = 576 + 4900 = 5476 = (x² + y²)²
So you now have:
x² - y² = 24
x² + y² = ±√5476 = ±74
…and from there it is trivial to solve.

Alternatively, x² + y² + 2xy = (x + y)² = ±74 + 70 = -4 or 144
So you can directly get = x + y, without having to solve for x or y themselves.

[yukigo]

y = 35/x
x² - y² = 24
x² - (35/x)² = 24
x⁴ - 24x² - 35² = 0
x⁴ - 24x² - 5²*7² = 0
(x² + 5²) (x² - 7²) = 0
x² = -5²(undefined), 7²
1. x = 7, y = 35/7 = 5 ==> x + y = 12
2. x = -7, y = 35/(-7) = -5 ==> x + y = -12
Ans: x + y = ±12

[mohadelassi]

This type of equation does not need the extensive work. The answer is 5+7 =12
7^2 - 5^2= 24

[julie.isbill]

Wouldn't it be much simpler to use unique prime factorization, which gives us x = +-7 and y = +-5 instantly ( xy = 35, so one is +-5 and the other is +-7, equation (1) tells us which is which). So we get x + y = +-12. Similar arguments hold for the two complex solutions x = +-5i and y = -+7i. Note the alternating + and - in this case. This gives x + y = +-2i.

[lcarliner]

The quickest and easiest way to solve is to factor the right-hand value. The only solution prime numbers 7 and 5! Assuming that the values for x equal 7 and Y equals 5, substituting these into the first equation confirm this!

[mingzhong]

let a = x^2 and b = y^2, so a-b = 24 and a*b = 35^2. By eliminating b, we have a(a-24) = 35*2, so a^2-24a-35*2 = 0. It is easily to know a = (24+/-sqrt (24^2+4*35^2))/2, so a = 49 or -25 (which can be excluded because a = x^2 cannot be negative). b = a-24 =25. It is easily to find either x=7 and y =5 or x=-7 and y = -5.

[rijwanmohammed]

Take all the factors of 24 i.e., 1x24, 2x12, 3x8, 4x6 or in opposite order like 24x1, 12x2 etc..and eqaute it to (x-y)(x+y) you'll find that when x+y is 12 only its will satisfy positive value of 35 for xy.

[BigKing]

take the second equation: factors of 35 is 1*35 and 7*5.Firstly 1 * 35 doesn't satisfy the first equation but 7 *5 is a possibility. The answer to the first equation is a positive number. Since 7 is the bigger number that means x is 7 and y is 5. (7 + 5 = 12).

[subhashdeshpande]

Very labouriously solved ! We call this as Hamali !!

[satrajitghosh]

(x-y)(x+y) = 24
case I : x-y = 1, x+y = 24
i.e x = 12+ 1/2 and y = 11+ 1/2
does not satisfy xy = 35
case II : x-y = 2, x+y = 12
i.e x = 7, y = 5
satisfies xy = 35
case III : x-y = 3, x+y = 8
i.e x = 5+1/2, y = 2+ 1/2
does not satisfy xy = 35
case III : x-y = 4, x+y = 6
i.e x = 5, y = 1
does not satisfy xy = 35
Hereby x= 7, y = 5 is the only feasible solution

[balaji]

For those of you commenting that you were able to get the answer 12, he is also providing complex number answer which you guys are not providing. Since the question did not specify explicitly whether the answer is in real or integer or complex, you have to provide all.