maths question? i need to know how to solve this simultaneusly?

Question

the problem asks for me to solve the following equation for X and Y simultaneusly.

xy
9^ = 81

x-y
3^ = 9

Any help you could give me would be fantastic, thanks heaps.

9 (to the power of) xy = 81
3 (to the power of) x - y) = 9

Answer 1

Can you clarify the formulas. Is it

xy^9= 81 and x-y^3=9. I assume that by ^ you mean to the power of.

Okay if 9^(xy)=81 then xy=2 (ie 9^2=81)
if 3^(x-y)=9 then x-y=2 (ie 3^2=9)

Solving these equations
from the first y=2/x
Substituting into 2nd x-2/x=2x
multiply by x and rearrange and you get x^2-2x-2=0.

Sorry it has been too long and I can't remember the quadratic equation. But use to solve for x and then work out the figures for y. I expect that there will be a pair of results.

Hope this helps.

Answer 2

:: First part
9 (to the power of) xy = 81
9^xy = 81
(3²)^xy = 3^4
Working with exponents:
2 * xy = 4
xy = 4 : 2
xy = 2
To justify, substitute the variables xy for exponent 2.
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The equal way:
3 (to the power of) x - y) = 9^
3^(x - y) = 3²
x - y = 2
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:: Second part
Finding the value of x and y.
Equation I) xy = 2
Equation II) x - y = 2

Substituting:
I) x = 2/y

II) x - y = 2
2/y - y = 2
2- y² - 2y = 0 (-1)
Baskhara: y² + 2y -2 = 0 ==> a = 1; b = 2; c = -2
delta = b² - 4ac
d = 2² - 4*1*-2
d = 4 + 8
d = 12

y = (-b +/- \/delta) : 2a
y = (-2 +/- \/12) : 2*1
y = (-2 + 2\/3) : 2
y' = (-2/2) + (2*1.732/2)
y' = -1 + 1.732 = 0.732 round
y"= -1 - 1.732 = -2.732 round

x - y' = 2
x - 0.732 = 2
x = 2 + 0.732
x' = 2.732

x - y" = 2
x - (-2.732) = 2
x = 2 - 2.732
x" = -0.732

Therefore:
x = -0.732 or 2.732
y = -2.732 or .732
I hope to be helped you!
::x::

Answer 3

9 = 3^2
9^xy =(3^2)^xy
ie 3^2xy =81 =3^4
2xy=4
xy=2
and 3^(x-y) = 9 =3^2
x-y =2
when bases are equal,their exponents must be equal
(x+y)^2 = x^2 + y^2 +2xy
(x-y)^2 = x^2 + y^2 - 2xy
(x+y)^2 = (x-y)^2 + 4xy
x+y = sqrt[(x-y)^2 + 4xy]
hence x+y = sqrt[4+4 * 2] =sqrt[12]
=2*sqrt(3)
x-y=2
x+y=sqrt(12)
on adding the above equations you get
2x = 2 + sqrt (12)
x = 1+ sqrt(12)/2
x= 1+sqrt3
x = 1+1.732
x=2.732
x-y = 2
implies y =x-2
therefore y =2.732-2
y=0.732
hence x = 2.732 and y = 0.732

Answer 4

9^(xy) = 81, so obviously xy = 2.
3^(x-y) = 9, so obviously x-y = 2.

From the second equation: x = y + 2.
Plugging into the first: (y+2)*y = 2
y^2 + 2y = 2
y^2 + 2y - 2 = 0
Using the quadratic formula:
y = (-2 (+/-) sqrt(4 + 8))/2
y = -1 (+/-) sqrt(3)
y = 0.732 or y = -2.732
... and x = y+2.

There are two solutions:
x = 2.732, y = 0.732
x = -0.732, y = -2.732.

Hope that helps!

Answer 5

9^(xy) = 81
9^(xy)= 9^2
Therefore xy=2 [If the bases are equal the exponents are also equal
3^(x-y)=9
3^(x-y)=3^2
Therefore x-y=2
Now you have two simultaneous equations:
xy=2
x-y=2
Substitute from the first one (y=2/x) in the second one
x-2/x=2
multiply both sides by x
x^2-2x-2=0
Solve this equation you get x=1+sqroot(3)=2.732
or x = 1-sqroot(3) = -0.732
and substituting these values in the first simultaneous eqn. you get y= sqroot(3)=0.732 or y=-1-sqroot(3)=-2.732
The final answer is x= 1+sqroot(3), y=sqroot(3)
or x=1-sqroot(3), y=-1-sqroot(3)

Answer 6

re type ur questioin by inserting parentheses where appropraite and u sure will get help