neolmar4144 neolmar4144
  • 21-10-2019
  • Mathematics
contestada

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

Respuesta :

nuuk nuuk
  • 22-10-2019

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

Step-by-step explanation:

Given

[tex]2^{2x+y-1}=32[/tex]

[tex]2^{2x+y-1}=2^5[/tex]

thus on comparing  

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

[tex]4^{x-2y}=2[/tex]

[tex]2^{2(x-2y)}=2[/tex]

On comparing exponent of 2

[tex]x-2y=\frac{1}{2}[/tex] ------2

we get x=2.5

substitute x in 1

y=1

(x,y)=(2.5,1)

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