solve for x and y , the simultaneous equations 4^x+3 = 32(2^x+y) and 9^x + 3^y = 10

Answer 1

Assuming the usual carelessness with parentheses, I get

4^(x+3) = 32(2^(x+y))
2^(2(x+3)) = 2^5*2^(x+y)
2x+6 = 5+x+y
x = y-1

9^(y-1) + 3^y = 10
3^(2y-2) + 3^y = 10
3^(2y)/9 + 3^y = 10
3^(2y) + 9*3^y - 90 = 0
(3^y + 15)(3^y - 6) = 0
3^y = 6
y = log6/log3
x = y-1

That answer is unusual, so check to make sure I got things right.

Answer 2

Wrong answer🙄

Answer 3

Right answer. The answer is:
y = 1.6309, x = 0.6309

Answer 4

How did u get (3^y + 15)(3^y - 6) = 0 ?

Answer 5

I didn't understand how did you get (3^y+15) (3^y-6)

Answer 6

9^x + 3^y =10. [x=y-1]
3^2x + 3^y =10
3^(2y-2) + 3^y =10
3^2y / 3^2 + 3^y = 10
3^2y + (3^2)(3^y) =10
Here assume 3^y=u
u² + 9u -10=0
Using quadratic function
u=6 or -15
(Skip -ve value )

u=6. 3^y=6
log base 3 of 6= y

Change base value
lg6/lg3 =y
y=1.63
x=0.63