Question

The solution of the system of equations \[\dfrac{{2x + 5y}}{{xy}} = 6\] and \[\dfrac{{4x - 5y}}{{xy}} + 3 = 0\] (where \[x \ne 0,y \ne 0\]) respectively is- …………………………………….
A. \[1,2\]
B. \[0,0\]
C. \[ - 1,2\]
D. \[1, - 2\]

Answer 1

Hint: Here, we have to find the solution for the given system of equations. First, we will convert the system of equations into a linear equation of two variables by substitution. Then we will solve the linear equation by elimination method to find the values of the given variable. A linear equation is an equation of the highest order as 1 with two variables.

Complete Step by Step Solution:
We are given the system of equations \[\dfrac{{2x + 5y}}{{xy}} = 6\] and \[\dfrac{{4x - 5y}}{{xy}} + 3 = 0\]
Considering, the first equation, we have
\[\dfrac{{2x + 5y}}{{xy}} = 6\]
By cross-multiplying, we get
\[ \Rightarrow 2x + 5y = 6xy\]
Dividing by \[xy\]on both the sides, we get
\[ \Rightarrow \dfrac{{2x}}{{xy}} + \dfrac{{5y}}{{xy}} = \dfrac{{6xy}}{{xy}}\]
By cancelling the similar terms, we get
\[ \Rightarrow \dfrac{2}{y} + \dfrac{5}{x} = 6\] ………………………………………………………..\[\left( 1 \right)\]
Now, considering the second equation, we have
\[\dfrac{{4x - 5y}}{{xy}} + 3 = 0\]
By cross-multiplying, we get
\[ \Rightarrow \dfrac{{4x - 5y}}{{xy}} + 3 \times \dfrac{{xy}}{{xy}} = 0\]
By rewriting the equation, we get
\[ \Rightarrow 4x - 5y + 3xy = 0\]
\[ \Rightarrow 4x - 5y = - 3xy\]
Dividing by \[xy\] on both the sides, we get
\[ \Rightarrow \dfrac{{4x}}{{xy}} - \dfrac{{5y}}{{xy}} = \dfrac{{ - 3xy}}{{xy}}\]
By cancelling the similar terms, we get
\[ \Rightarrow \dfrac{4}{y} - \dfrac{5}{x} = - 3\]……………………………………………………………\[\left( 2 \right)\]
Now, Substituting \[\dfrac{1}{x} = a\] and \[\dfrac{1}{y} = b\] in equation \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[2b + 5a = 6\]……………………………………………………………\[\left( 3 \right)\]
\[4b - 5a = - 3\] …………………………………………………………\[\left( 4 \right)\]
Now, we have to solve the equations \[\left( 3 \right)\]and \[\left( 4 \right)\], to find the value of \[a,b\]
By adding equations \[\left( 3 \right)\] and \[\left( 4 \right)\], we get
\[6b + 0a = 3\]
\[ \Rightarrow 6b = 3\]
Dividing both side by 6, we get
\[ \Rightarrow b = \dfrac{3}{6}\]
\[ \Rightarrow b = \dfrac{1}{2}\]
Substituting \[b = \dfrac{1}{2}\] in equation \[\left( 3 \right)\], we get
\[2\left( {\dfrac{1}{2}} \right) + 5a = 6\]
Multiplying the terms, we get
\[ \Rightarrow 1 + 5a = 6\]
Subtracting 1 from both the sides, we get
\[ \Rightarrow 5a = 6 - 1\]
\[ \Rightarrow 5a = 5\]
Dividing both side by 5, we get
\[ \Rightarrow a = \dfrac{5}{5}\]
\[ \Rightarrow a = 1\]
Now we will substitute the value of \[a\] and \[b\] in the equation \[x = \dfrac{1}{a}\] and \[y = \dfrac{1}{b}\]. Therefore, we get
When \[a = 1\] then
\[x = \dfrac{1}{1} = 1\]
When \[b = \dfrac{1}{2}\] then
\[y = \dfrac{1}{{\dfrac{1}{2}}} = 2\]
Thus, the solution is\[\left( {1,2} \right)\].

Therefore, the solution of the system of equations \[\dfrac{{2x + 5y}}{{xy}} = 6\] and \[\dfrac{{4x - 5y}}{{xy}} + 3 = 0\] is \[\left( {1,2} \right)\].
Thus, Option A is correct.

Note:
We know that the linear equation of two variables should have constant terms to solve the equation. If the constant term has variables, then the variable has to be removed to solve the equation. The variables can be removed by substitution. But we should remember that if the variables are substituted for some variable, then it has to be substituted again to find the solution for the given variables.