Re: Find the value of x & y

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• Subject: [mg127832] Re: Find the value of x & y
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Sat, 25 Aug 2012 04:26:21 -0400 (EDT)
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• References: <20120824090532.3FD59684A@smc.vnet.net>

```eqn = 16^(x*x + y) + 16^(x + y*y) == 1;

If you guess from the symmetry that perhaps x == y

sol = Solve[{eqn, x == y}, {x, y}, Reals][[1]]

{x -> -(1/2), y -> -(1/2)}

Verifying that assumption,

eqn /. sol

True

>From this assumption the solution can be arrived at any number of ways.

Reduce[{eqn, x == y}, {x, y}, Reals] // ToRules

{x -> -(1/2), y -> -(1/2)}

FindInstance[{eqn, x == y}, {x, y}, Reals][[1]]

{x -> -(1/2), y -> -(1/2)}

FindRoot[{eqn, x == y}, {{x, 1}, {y, 1}}]

{x -> -0.5, y -> -0.5}

Bob Hanlon

On Fri, Aug 24, 2012 at 5:05 AM,  <mainakju05 at gmail.com> wrote:
> Find the value of x & y. We have only one equation: 16x.x+y + 16x+y.y = 1(16 to the power x.x + y + 16 to the power x+y.y = 1.
>
> I have found that X =-1/2 & y = -1/2 satisfying this equation. But I got it through trial method. It will be helpful, if anyone can solve this.
>

```

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