Question

1/2x+1/3y=2,1/3x+1/2y=13/6 (xand y is not =0)​

Answer 1

You can answer this question by substitution or elimination method.

I am solving this by substitution..

1/2x=2-1/3y

1/2x=(6y-1)/3y

x=(6y-1)/3y×2

x=(12y-2)/3y

putting this value of x in the second equation

we will get the value of y

and when we will put the value of y in the first equation we will get the value of x

then you will end up with the values of x and y.

Hope it helps ....

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Answer 2

Answer:

(x,y)=($\frac{1}{2}, \frac{1}{3}$)

Step-by-step explanation:

$\frac{1}{2x} +\frac{1}{3y} =2-----(1)\\\\\frac{1}{3x} +\frac{1}{2y}=\frac{13}{6}------(2)$

Multiplying (1) with 12 and equation (2) with 18

$\frac{1}{2x} +\frac{1}{3y} =2-----(1) *12\\\\\frac{1}{3x} +\frac{1}{2y}=\frac{13}{6}------(2) *18$

$\frac{6}{x}+ \frac{4}{y}=24$

$\frac{6}{x}+ \frac{9}{y}=39$

subtracting equation 1) from (2)

$\frac{6}{x} +\frac{9}{y} -\frac{6}{x} -\frac{4}{y} =39-24$

$\frac{9}{y} -\frac{4}{y} =15$

$\frac{1}{y} =3\\\\y=\frac{1}{3}$

Substituting y value in equation (1)

$\frac{6}{x} +4*3=24$

$\frac{1}{x} =2\\\\x=\frac{1}{2}$

( x and y is not equal to 0)

(x,y)= $(\frac{1}{2}, \frac{1}{3} )$

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