Question

Solve this system algebraically.

Answer 1

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.

Answer 2

The answer for this question is the last option which is D.