JAMB Mathematics 2015 Paper

Concept:Use the laws of indices by rewriting all powers as base 3, then solve the resulting simultaneous equations.Explanation:Given: $\dfrac{27^x}{81^{x+2y}} = 9$ and $x + 4y = 0$.Rewrite each number as a power of 3: $27 = 3^3$, $81 = 3^4$, and $9 = 3^2$.So, $27^x = (3^3)^x = 3^{3x}$ and $81^{x+2y} = (3^4)^{x+2y} = 3^{4x+8y}$.The first equation becomes:$$\frac{3^{3x}}{3^{4x+8y}} = 3^2$$Using the quotient rule for indices, subtract exponents:$$3^{3x - (4x + 8y)} = 3^2$$$$3^{-x - 8y} = 3^2$$Since the bases are equal, equate the exponents:$$-x - 8y = 2 \quad \text{...(1)}$$The second equation is:$$x + 4y = 0 \quad \text{...(2)}$$Add equation (1) and equation (2):$$(-x + x) + (-8y + 4y) = 2 + 0$$$$-4y = 2$$$$y = -\frac{1}{2}$$Substitute $y = -\frac{1}{2}$ into equation (2):$x + 4\left(-\frac{1}{2}\right) = 0$$x - 2 = 0$$$x = 2$$Answer:$x = 2$ and $y = -\frac{1}{2}$.This corresponds to option B.