Question

Solve the following systems of equations. 2^(2x+y−1) = 32 4^(x−2y) = 2

Answer 1

Answer:(x,y)=(2.5,1)

Step-by-step explanation:

Given

$2^{2x+y-1}=32$

$2^{2x+y-1}=2^5$

thus on comparing  

2x+y=6------1

$4^{x-2y}=2$

$2^{2(x-2y)}=2$

On comparing exponent of 2

$x-2y=\frac{1}{2}$ ------2

we get x=2.5

substitute x in 1

y=1

(x,y)=(2.5,1)