Question

Solve the following system of equations by the elimination method.
\[8x-3y=5xy\], \[6x-5y=-2xy\], \[x\ne 0\], \[y\ne 0\].

Answer 1

Hint:First of all divide both the equation by xy and substitute \[\dfrac{1}{y}=A\] and \[\dfrac{1}{x}=B\]. Now, equate the coefficient of B in both the equations and subtract the second equation from first to eliminate the terms containing B and find the value of A. Now from the values of A and B, find the values of x and y.

Complete step-by-step answer:
In this question, we have to solve the equations 8x – 3y = 5xy and 6x – 5y = – 2xy by elimination method. First of all, let us consider the equations given to us.
\[8x-3y=5xy....\left( i \right)\]
\[6x-5y=-2xy....\left( ii \right)\]
First of all, let us divide equation (i) and (ii) by xy, we get,
\[\dfrac{8x}{xy}-\dfrac{3y}{xy}=\dfrac{5xy}{xy}\]
\[\dfrac{8}{y}-\dfrac{3}{x}=5....\left( iii \right)\]
And, \[\dfrac{6x}{xy}-\dfrac{5y}{xy}=\dfrac{-2xy}{xy}\]
\[\dfrac{6}{y}-\dfrac{5}{x}=-2...\left( iv \right)\]
Now by substituting \[\dfrac{1}{y}=A\] and \[\dfrac{1}{x}=B\] in equation (iii) and (iv), we get,
\[8A-3B=5....\left( v \right)\]
\[6A-5B=-2....\left( vi \right)\]
Now, to use the elimination method, we have to equate the coefficient of either A or B in the above equation. So by multiplying by 5 in equation (v), we get,
\[40A-15B=25....\left( vii \right)\]
Now by multiplying by 3 in equation (vi), we get,
\[18A-15B=-6....\left( viii \right)\]
Now, by substituting equation (viii) from equation (vii), we get,
\[\begin{align}
  & 40A-15B=25 \\
 & \underline{\begin{align}
  & 18A-15B=-6 \\
 & \text{ + +} \\
\end{align}} \\
 & \text{ }22A=31 \\
\end{align}\]
So, we get, \[A=\dfrac{31}{22}\]
By substituting \[A=\dfrac{31}{22}\] in equation (v), we get,
\[8\left( \dfrac{31}{22} \right)-3B=5\]
\[3B=8\left( \dfrac{31}{22} \right)-5\]
\[3B=\dfrac{4\left( 31 \right)}{11}-5\]
\[3B=\dfrac{124-55}{11}=\dfrac{69}{11}\]
So, we get, \[B=\dfrac{23}{11}\].
Now, we know that \[A=\dfrac{1}{y}=\dfrac{31}{22}\]
So, we get \[y=\dfrac{22}{31}\]
Also, \[B=\dfrac{1}{x}=\dfrac{23}{11}\]
So, we get, \[x=\dfrac{11}{23}\]
Hence, we have found the values of x and y that is \[x=\dfrac{11}{23}\] and \[y=\dfrac{22}{31}\] by the elimination method.
Therefore option (a) is the correct answer.

Note: In these types of questions, students are advised to substitute \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] as some other variable to avoid the confusion. Also, don’t forget to convert these variables back to x and y. Students should always cross-check their answers by substituting the values of x and y in the initial equations and checking if they are satisfying the equation or not.