Note that each of the bases here are powers of 3:
3¹ = 3
3² = 9
3³ = 27
3⁴ = 81
3⁵ = 243
So the given equations can be rewritten as
[tex]\dfrac{81^{x+2y}}{27^{x-y}} = \dfrac{\left(3^4\right)^{x+2y}}{\left(3^3\right)^{x-y}} = \dfrac{3^{4x+8y}}{3^{3x-3y}} = 3^{x+11y} = 3^5[/tex]
and
[tex]\dfrac{243^{x+3y}}{81^{x+2y}} = \dfrac{\left(3^5\right)^{x+3y}}{\left(3^4\right)^{x+2y}} = \dfrac{3^{5x+15y}}{3^{4x+8y}} = 3^{x+7y} = 3^{-3}[/tex]
The bases on either side are the same, so the exponents must match and
x + 11y = 5
x + 7y = -3
Solve for x and y. Eliminate x and solve for y :
(x + 11y) - (x + 7y) = 5 - (-3)
(x - x) + (11y - 7y) = 5 + 3
4y = 8
y = 2
Solve for x :
x + 11•2 = 5
x + 22 = 5
x = -17
