Calculate the value of x and y if (27x ÷ 81x+2y = 9 ,x + 4y = 0

A.

x = 1, y = 1/2

B.

x = 2, y = – 1/2

C.

x − 0, y = 1

D.

x = 2, y = –1

Correct answer is B

\(27^x ÷ 81^{(x + 2y)} = 9 \\

(27)x = 9 × 81^{(x+2y)} \\

(3^3 )^x =32 \times 3^{4(x + 2y)} \\

=3^{(2 + 4x + 8y)}\\

3^{3x} = 3^{ (2 + 4x + 8y)}\\

3x = 2 + 4x + 8y\\

3x − 4x − 8y = 2 … … … (1)\\

x + 4y = 0 … … … (2)\\

− 4y = 2\\

y = (− 2) ÷ 4 = − ½\\

y = − ½\\ \)

Substitute the value of y into equation (2)

i.e x + 4y = 0

x + 4( − 1/2) = 0

x − 2 = 0

x = 2

∴ x = 2,y = − ½)

Method II

\( 27^x ÷ 31^{(x + 2y) }= 9\\

3^{3x} × 3^{( − 4x − 8y)} = 32\\

3^{(3x − 8y)} = 32\\

− x − 8y=2 ……… (1)\\

x + 4y = 0 ……… (2)\\

− 4 = 2\\

y= 2/4 = ½\\

y = ½ \)

Substitute the value of y into equation 2

x + 4y=0

x + 4 (− 1) ÷ 2) = 0

x − 2 = 0

x = 2

x = 2, y = ½