Answer: The correct option is
(D) \((x,y)=\left(\dfrac{40}{41},\dfrac{58}{41}\right).\)
Step-by-step explanation: We are given to solve the following system of equations algebraically:
\(7x-2y=4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\5y+3x=10~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\)
Multiplying equation (i) by 5, we have
\(5(7x-2y)==4\times 5\\\\\Rightarrow 35x-10y=20~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)\)
Multiplying equation (ii) by 2, we have
\(2(5y+3x)=2\times 10\\\\\Rightarrow 10y+6x=20~~~~~~~~~~~~~~~~~~~~~~~~(ii)\)
Adding equations (iii) and (iv), we get
\(35x+6x=20+20\\\\\Rightarrow 41x=40\\\\\Rightarrow x=\dfrac{40}{41}.\)
Substituting the value of x in equation (ii), we get
\(5y+3\times\dfrac{40}{41}=10\\\\\Rightarrow 5y+\dfrac{120}{41}=10\\\\\Rightarrow 5y=10-\dfrac{120}{41}\\\\\Rightarrow 5y=\dfrac{290}{41}\\\\\Rightarrow y=\dfrac{290}{205}\\\\\Rightarrow y=\dfrac{58}{41}.\)
Thus, the required solution is
\((x,y)=\left(\dfrac{40}{41},\dfrac{58}{41}\right).\)
Option (D) is correct.
