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,

& 40A-15B=25 \\\ & \underline{\begin{aligned} & 18A-15B=-6 \\\ & \text{ + +} \\\ \end{aligned}} \\\ & \text{ }22A=31 \\\ \end{aligned}$$ So, we get, $$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.