Question 40

Jambmaths question: 

If ${{2}^{x+y}}=16$ and ${{4}^{x-y}}=\frac{1}{32}$ , find the values of x and y 

Option A: 

$x=\tfrac{3}{4},\text{ }y=\tfrac{11}{4}$

Option B: 

$x=\tfrac{4}{3},\text{ }y=\tfrac{13}{4}$

Option C: 

$x=\tfrac{2}{3},\text{ }y=\tfrac{4}{5}$

Option D: 

$x=\tfrac{2}{3},\text{ }y=\tfrac{13}{4}$

Jamb Maths Solution: 

$\begin{align}  & {{2}^{x+y}}=16 \\ & {{2}^{x+y}}={{2}^{4}} \\ & x+y=4---(i) \\ & {{4}^{x-y}}=\frac{1}{32} \\ & {{2}^{2(x-y)}}={{2}^{-5}} \\ & 2x-2y=-5---(ii) \\ &  \\ & x+y=4---(i) \\ & 2x-2y=-5---(ii) \\ & \text{From (i) }x=4-y \\ & \text{Substitute }4-y\text{ for }x\text{ into equation }(ii) \\ & 2(4-y)-2y=-5 \\ & 8-2y-2y=-5 \\ & 8-4y=-5 \\ & 4y=8+5 \\ & y=\frac{13}{4} \\ & \text{substitute }\frac{13}{4}\text{ for }y\text{ in }x=\text{4}-y \\ & x=4-\frac{13}{4}=\frac{3}{4} \\ & x=\frac{3}{4},\text{ }y=\frac{13}{4} \\\end{align}$

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