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The pair of equations \[{3^{x + y}} = 81\], \[{81^{x - y}} = 3\] has
A.No solution
B.The solution \[x = 2\dfrac{1}{2}\],\[y = 2\dfrac{1}{2}\]
C.The solution \[x = 2\], \[y = 2\]
D.The solution \[x = 2\dfrac{1}{8}\],\[y = 1\dfrac{7}{8}\]

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Last updated date: 17th Jun 2024
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Answer
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Hint: First we will first rewrite the number 81 into powers of 3 in the equation and then use the power rule that if \[{a^x} = {a^y}\], then \[x = y\]. Then we will simplify the equations to find the value of \[x\] and \[y\].

Complete step-by-step answer:
We are given that the pair of linear equations
\[{3^{x + y}} = 81{\text{ ......eq.(1)}}\]
 \[{81^{x - y}} = 3{\text{ ......eq(2)}}\]
Rewriting the number 81 into powers of 3 in the equation (1), we get
\[ \Rightarrow {3^{x + y}} = {3^4}\]
Using the power rule that if \[{a^x} = {a^y}\], then \[x = y\] in the above equation, we get
\[ \Rightarrow x + y = 4\]
Subtracting the above equation by \[x\] on both sides, we get
\[
   \Rightarrow x + y - x = 4 - x \\
   \Rightarrow y = 4 - x{\text{ .......eq.(3)}} \\
 \]
Rewriting the number 81 into powers of 3 in the equation (2), we get
\[
   \Rightarrow {3^{4\left( {x - y} \right)}} = 3 \\
   \Rightarrow {3^{4x - 4y}} = {3^1} \\
 \]
Using the power rule that if \[{a^x} = {a^y}\], then \[x = y\] in the above equation, we get
\[ \Rightarrow 4x - 4y = 1\]
Substituting the value of \[y\] from equation (3) in the above equation, we get
\[
   \Rightarrow 4x - 4\left( {4 - x} \right) = 1 \\
   \Rightarrow 4x - 16 + 4x = 1 \\
   \Rightarrow 8x - 16 = 1 \\
 \]
Adding the above equation by 16 on both sides, we get
\[
   \Rightarrow 8x - 16 + 16 = 1 + 16 \\
   \Rightarrow 8x = 17 \\
 \]

Dividing the above equation by 8 on both sides, we get
\[
   \Rightarrow \dfrac{{8x}}{8} = \dfrac{{17}}{8} \\
   \Rightarrow x = \dfrac{{17}}{8} \\
 \]
Substituting the value of \[x\] in the equation (3), we get
\[
   \Rightarrow y = 4 - \dfrac{{17}}{8} \\
   \Rightarrow y = \dfrac{{32 - 17}}{8} \\
   \Rightarrow y = \dfrac{{15}}{8} \\
 \]
Writing the value of \[x\] and \[y\] into the mixed fraction, \[{\text{Quotient}}\dfrac{{{\text{Remainder}}}}{{{\text{Divisor}}}}\] to match with the options, we get
\[ \Rightarrow x = 2\dfrac{1}{8}\]
\[ \Rightarrow y = 1\dfrac{7}{8}\]

Hence, option D is correct.

Note: We can avoid the final steps of mixed form by matching the denominators with the options and finding the correct one. We know that a linear system of two equations with two variables is any system that can be written in the form. A solution to a system of equations is a value of \[x\] and a value of \[y\] that, when substituted into the equations, satisfies both equations at the same time.