Solve the following simultaneous equations :
\[\dfrac{1}{x} + \dfrac{1}{y} = 8\] ; \[\dfrac{4}{x} - \dfrac{2}{y} = 2\]
\[\left( A \right)\] \[x = \dfrac{1}{3}\] , \[y = \dfrac{1}{5}\]
\[\left( B \right)\] \[x = \dfrac{1}{2}\] , \[y = \dfrac{1}{5}\]
\[\left( C \right)\] \[x = \dfrac{1}{3}\] , \[y = \dfrac{1}{7}\]
\[\left( D \right)\] \[x = \dfrac{1}{2}\] , \[y = \dfrac{1}{7}\]
Answer
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Hint: We have to find the value of both the variables x and y from the given expression \[\dfrac{1}{x} + \dfrac{1}{y} = 8\] ; \[\dfrac{4}{x} - \dfrac{2}{y} = 2\] . We solve this question using the concept of solving the linear equations . We should also have the knowledge of the concept of solving the equation using the elimination method . First we will consider the relation of \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] with another variable and then simplify the relation such that we obtain two linear equations . Then we will solve the two linear equations using the elimination method . And hence find the value of the variable \[x\] and \[y\] .
Complete step by step answer:
Given, \[\dfrac{1}{x} + \dfrac{1}{y} = 8\] , \[\dfrac{4}{x} - \dfrac{2}{y} = 2\]
Let us consider that the relation \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] are such that as given
\[\;\dfrac{1}{x} = a\] and \[\;\dfrac{1}{y} = b\]
Substituting the values of the relations , we can write the expressions for linear equations as
\[a + b = 8 - - - \left( 1 \right)\]
\[4a - 2b = 2 - - - \left( 2 \right)\]
Now we will solve two linear expressions using the elimination method.
Multiplying equation \[\left( 1 \right)\] by \[4\] , we can write the expression as
\[4 \times \left[ {a + b = 8} \right]\]
\[4a + 4b = 32 - - - \left( 3 \right)\]
Subtracting equation \[\left( 3 \right)\] from equation \[\left( 2 \right)\] , we get the value of $ b$ as
\[4a + 4b - \left[ {4a - 2b} \right] = 32 - 2\]
\[4a + 4b - 4a + 2b = 30\]
On solving , we can write the expression as
\[6b = 30\]
\[b = 5\]
Multiplying equation \[\left( 1 \right)\] by \[2\] , we can write the expression as
\[2 \times \left[ {a + b = 8} \right]\]
\[2a + 2b = 16 - - \left( 4 \right)\]
Adding both equation \[\left( 4 \right)\] and equation \[\left( 2 \right)\] , we get the value of a as
\[2a + 2b + \left[ {4a - 2b} \right] = 16 + 2\]
\[2a + 2b + 4a - 2b = 18\]
On solving , we can write the expression as
\[6a = 18\]
\[a = 3\]
Substituting the values of a and b back into the relations , we get the values as
\[x = \dfrac{1}{3}\] and \[y = \dfrac{1}{5}\]
Hence, we get the value of the variables \[x\] and \[y\] as \[\dfrac{1}{3}\] and \[\dfrac{1}{5}\] respectively.
Thus, the correct option is option \[\left( A \right)\].
Note:
For the value of the functional expression, we can use any of the methods of solving the equation. We could also use the substitution method or the cross multiplication method to solve the value of the functional expression. But we used the elimination method as it is easier and less complicated then the other two methods.
Complete step by step answer:
Given, \[\dfrac{1}{x} + \dfrac{1}{y} = 8\] , \[\dfrac{4}{x} - \dfrac{2}{y} = 2\]
Let us consider that the relation \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] are such that as given
\[\;\dfrac{1}{x} = a\] and \[\;\dfrac{1}{y} = b\]
Substituting the values of the relations , we can write the expressions for linear equations as
\[a + b = 8 - - - \left( 1 \right)\]
\[4a - 2b = 2 - - - \left( 2 \right)\]
Now we will solve two linear expressions using the elimination method.
Multiplying equation \[\left( 1 \right)\] by \[4\] , we can write the expression as
\[4 \times \left[ {a + b = 8} \right]\]
\[4a + 4b = 32 - - - \left( 3 \right)\]
Subtracting equation \[\left( 3 \right)\] from equation \[\left( 2 \right)\] , we get the value of $ b$ as
\[4a + 4b - \left[ {4a - 2b} \right] = 32 - 2\]
\[4a + 4b - 4a + 2b = 30\]
On solving , we can write the expression as
\[6b = 30\]
\[b = 5\]
Multiplying equation \[\left( 1 \right)\] by \[2\] , we can write the expression as
\[2 \times \left[ {a + b = 8} \right]\]
\[2a + 2b = 16 - - \left( 4 \right)\]
Adding both equation \[\left( 4 \right)\] and equation \[\left( 2 \right)\] , we get the value of a as
\[2a + 2b + \left[ {4a - 2b} \right] = 16 + 2\]
\[2a + 2b + 4a - 2b = 18\]
On solving , we can write the expression as
\[6a = 18\]
\[a = 3\]
Substituting the values of a and b back into the relations , we get the values as
\[x = \dfrac{1}{3}\] and \[y = \dfrac{1}{5}\]
Hence, we get the value of the variables \[x\] and \[y\] as \[\dfrac{1}{3}\] and \[\dfrac{1}{5}\] respectively.
Thus, the correct option is option \[\left( A \right)\].
Note:
For the value of the functional expression, we can use any of the methods of solving the equation. We could also use the substitution method or the cross multiplication method to solve the value of the functional expression. But we used the elimination method as it is easier and less complicated then the other two methods.
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