Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 17th Jun 2024
Total views: 403.2k
Views today: 5.03k
Verified
403.2k+ views
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 $y$.

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.