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Thursday, January 31, 2008

Simple Algebra

A reader of this page asks:

"If x²+y² = 36 and (x+y)² = 64, what is the value of x∙y?"

Help a fellow puzzle fan discover the values of x and y.

22 comments:

Anonymous said...

how can this be possible? i've gone through so many combinations. i've gotten quite far, but at the last minute something turns out wrong. i don't want the answer, i just want to know whether or not it's actually possible.

Erik said...

It is possible, but the values of x and y are not integers.

Riccardo Antonelli said...

The solution of the system gives four possible values for x and y, and for all of them xy is 14.

Riccardo Antonelli (2) said...

aargh, never mind, the product of the solutions is c/b (c and b are the third and second degrees coefficients) so it is simply 14. I've done many unuseful calculations.

Anonymous said...

The problem is simple. Expand (x+y)squared and replace the value of Xsquared + Ysquared with 36.

we will arrive at 2xy = 64 -36

which gives xy =14

M said...

It's obvious that xy=14, but that still does not answer the original question : what is y and what is x?

elmer said...

The question "what is the value of x*y?" has been answered already and it is 14.

If you had been wondering what is x and y, there are two possible set of answers.

#1.
x = 5.414213562
y = 2.585786438

#2.
x = 2.585786438
y = 5.414213562

x and y values were determined using the quadratic equation formula.

Anonymous said...

64=(x+y)²=x²+y²+2xy
since x²+y²=36
2xy=64-36=28

(x-y)²=x²+y²-2xy=36-28=8

then x-y=2V2 or -2V2
(v=square root)

Naboshad said...

there is no answer. X2+Y2=36 is a circle from the origin with the rediuce of 6.
(x+y)2=64 is two lines which pass A( 0,8 ), B ( 8,0) and C( 0,-8), D( -8 ,0 ). these lines and that circle have no intercetion.

Anonymous said...

1. x2+y2= 36
2.(x+y)2= (x+y)(x+y)= x2+y2+2xy=64
3. 36 + 2xy = 64
4. 2xy=28
5.xy = 14

Anonymous said...

1. x2+y2= 36
2.(x+y)2= (x+y)(x+y)= x2+y2+2xy=64
3. 36 + 2xy = 64
4. 2xy=28
5.xy = 14

Anonymous said...

astronomical!!!ever!!

Anonymous said...

(x+y)^2=64
x+y=8

(x+y)^2=64
x^2+2xy+y^2=64
x^2+y^2=36
2xy=28
xy=14

x+y=8
xy=14

x+y=8
0x+y=14/x

x=8-14/x
x^2-8x+14=0
quadratic equation

V is square root symbol

x=4+-V2 about 5.4
y= about 2.6

Anonymous said...

You are all wrong. Start with (x+y)^2 = 64, so x+y = 8

Then look at the equation (x^2+y^2)= 36.

We know that the two numbers x + y = 8 so if we look at 5 and 3 we get 25 + 9 = 36

so either x or y = 3 or 5

so the answer is not 14 its actually 15.

Hope you all agree its simple really

Louie said...

this is really easy. so x is 3 and y is 5. 3 to the 2nd power is 9 5 to the second power is 25 25+9=36. Then, x+y or 3+5 = 8 to the second power is 64 so x is 3 and y is 5

Louie said...

to put things simpiliar x^2 + y^2 = 36.

x = 3
y = 5
3^2 = 9
5^2 = 25
25+9=36



(x+y)^2 = 64
(3+5) ^2 = 64
(8) ^2 = 64
8x8=64

So 3x5 = 15 not what you others are saying as 14. It is not 7 and 2

Anonymous said...

@Louie:

25+9 is NOT EQUAL to 36.
25+9=34

so your solution is incorrect. :-)

Anonymous said...

http://www05.wolframalpha.com/input/?i=solve+{x%C2%B2%2By%C2%B2+%3D+36%2C+(x%2By)%C2%B2+%3D+64}

fishavenue said...

Somebody was on the right track earlier. One of the equations is a circle with radius 6 centered at origin. The other equation yields two straight lines with slope -1. Both of these lines intersect the circle, so there are four solutions. The distance of these lines from the origin is 5.6569, or 4*sqrt(2), so these lines do indeed intersect the circle. The y intercept of one line is +8 and the y intercept of the second line is -8

abishek said...

simple: 64=x*x+y*y+2xy
but x*x+y*y=36so
2xy=64-36
xy=28/2
xy=14

Anonymous said...

After 20 comments still no correct answer! I liked the people who thought 25 + 9 = 36 though. Anyway, here's my effort:

Several people have correctly established that xy = 14 (1)

From (x + y)^2 = 64 it is clear that x+y = 8 (2a) or x+y = -8 (2b)

subbing equation 1 into equation 2 (or vice versa) yields the equation:
x^2 - 8x + 14 = 0

solving this leads to the solutions x = 4 + sqrt(2) and
x = 4 - sqrt(2).
Subbing these values back in equation (2a) yields values for y.

Going back and doing the same thing with equations (1) and (2b) eventually leads to the four solutions:
x = 4 + sqrt(2), y = 4 - sqrt(2)
x = 4 - sqrt(2), y = 4 + sqrt(2)
x = -4 + sqrt(2), y = -4 - sqrt(2)
x = -4 - sqrt(2), y = -4 + sqrt(2)

The first 2 of these were stated as decimals further up the page.

Anonymous said...

These are all the solutions to these equations.