2^(3x+6y) = 2^7
Then equation 1 becomes simply,
3x+6y = 7
"What happens in the event horizon, stays in the event horizon" -- Bob Brown (grandpa Bob)
According to 1st post, the OP apparently was learning
how to solve "2 unknowns, 2 equations" problems.
I showed one possible solution to his 1st equation.
This was mainly to see if the OP understood.
I made up a similar case:
x=3, y=2
2^x * 5^y = 200 [1]
sqrt(12y - 15) = x [2]
Then used Wolfram (to "see"):
http://www.wolframalpha.com/input/?i...t(12*y-15)%3Dx
So no problems with a "proper" set of 2 equations.
I then tried the OP's 2 equations:
http://www.wolframalpha.com/input/?i...rt(12*y-7)%3Dx
No results...Wolfram gave up!
So I assumed there was something wrong
with the equations...
My apologies...
I'm just an imagination of your figment !
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
6x+12y = 14 [1]
12y - 7 = xx [2]
so
6x+ (xx+7) = 14
x=1 or x=-7
trying in [1], x=1
Answ: x=1, y=2/3
Wolfram Solution
Last edited by Bob Brown MSEE; 06-20-2017 at 05:03 AM.
"What happens in the event horizon, stays in the event horizon" -- Bob Brown (grandpa Bob)
I'm just an imagination of your figment !
I wouldn't go this way. Instead:
. . . . .3x + 6y = 7
. . . . .12y - 7 = x^2
Multiply the first equation by -2:
. . . . .-6x - 12y = -14
. . . . .12y - 7 = x^2
Add down to get:
. . . . .-6x - 7 = x^2 - 14
. . . . .0 = x^2 + 6x - 7
. . . . .0 = (x + 7)(x - 1)
...and so forth.
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