Question

The pair of equations \[{{\text{3}}^{{\text{x + y}}}}{\text{ = 81}}\], \[{\text{8}}{{\text{1}}^{{\text{x - y}}}}{\text{ = 3}}\]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 \[x = 2\dfrac{1}{8},y = 1\dfrac{7}{8}\]

Answer 1

Hint: Since the given equation is in exponents, therefore, we’ll we simplify in such a way that the base of the left-hand side will be equal to the right-hand side. Then on equating the exponents we’ll get a pair linear equations in x and y. Then we solve those linear equations using the substitution method and will get the required answer.

Complete step by step Answer:

From the first equation, \[{{\text{3}}^{{\text{x + y}}}}{\text{ = 81}}\], we get,
\[ \Rightarrow {{\text{3}}^{{\text{x + y}}}}{\text{ = }}{{\text{3}}^{\text{4}}}\]
Now, we will use the fact that if \[{{\text{a}}^{\text{x}}}{\text{ = }}{{\text{a}}^{\text{y}}}\], then \[{\text{x = y}}\]
So we have, \[{\text{x + y = 4}}\]….(1)

From the second equation,
\[
  {\text{8}}{{\text{1}}^{{\text{x - y}}}}{\text{ = 3}} \\
   \Rightarrow {{\text{3}}^{{\text{4(x - y)}}}}{\text{ = 3}} \\
 \]
Now, we will use the fact that if \[{{\text{a}}^{\text{x}}}{\text{ = }}{{\text{a}}^{\text{y}}}\], then \[{\text{x = y}}\]
\[ \Rightarrow {\text{4(x - y) = 1}}\]

\[ \Rightarrow {\text{x - y = 0}}{\text{.25}}\]…….(2)

Now we have, \[{\text{x - y = 0}}{\text{.25}}\], \[{\text{x + y = 4}}\]
Taking \[{\text{x + y = 4}}\],
\[ \Rightarrow {\text{x = 4 - y}}\] (3)

Now substituting the value of x in \[{\text{x - y = 0}}{\text{.25}}\], we get,
\[ \Rightarrow {\text{4 - y - y = 0}}{\text{.25}}\]
\[ \Rightarrow 4{\text{ - 2y = 0}}{\text{.25}}\]
\[ \Rightarrow 3.75 = 2{\text{y}}\]
\[ \Rightarrow {\text{y}} = 1.875\]
Converting into a mixed fraction we get,

\[ \Rightarrow {\text{y = 1}}\dfrac{{\text{7}}}{{\text{8}}}\]

And now, substituting the value of y in (3), we get,
\[ \Rightarrow {\text{x}} = 4 - 1.875\]
\[ \Rightarrow {\text{x = 2}}{\text{.125}}\]
Now converting into a mixed fraction we get,
\[ \Rightarrow x = 2\dfrac{1}{8}\]

So, we have our solution as, \[x = 2\dfrac{1}{8}\]and \[{\text{y = 1}}\dfrac{{\text{7}}}{{\text{8}}}\], which is an option (d)

Note: Here to solve the pair of linear equations in two variables we have used the substitution method, we can also use the elimination method or the cross multiplication method to solve for the variables.

\[ \Rightarrow {\text{x + y = 4}}..........{\text{(i)}}\]
\[ \Rightarrow {\text{x - y = 0}}{\text{.25}}..........{\text{(ii)}}\]
On adding both equations
\[ \Rightarrow 2{\text{x = 4 + 0}}{\text{.25}}\]
\[ \Rightarrow 2{\text{x = 4}}{\text{.25}}\]
\[ \Rightarrow {\text{x = 2}}{\text{.125}}\]
Now converting into a mixed fraction we get,
\[\therefore x = 2\dfrac{1}{8}\]
Substituting the value of x in equation(i)

\[ \Rightarrow {\text{2}}{\text{.125 + y = 4}}\]
\[ \Rightarrow {\text{y = 4 - 2}}{\text{.125}}\]
\[\therefore {\text{y = 1}}{\text{.875}}\]
So, we have our solution as, \[x = 2\dfrac{1}{8}\]and \[{\text{y = 1}}\dfrac{{\text{7}}}{{\text{8}}}\].