Question

Solve linear equation 13+3/y+2/y=0, 2+5/x-4/y=0 by substitution method

Answer 1

Solution :

EQ 1. 13+3/x+2/y=0

EQ2. 2+5/x-4/y=0

from eq1. : x= -13-2/y

= -(13+2/y)

Now, put value of x in eq2. :

2+5/x-4/y=0

2+5/-13-2/y-4/y=0

2+5×-2/-13-4/y =0

2+-10/-13-4/y =0

2+10/13-4/y=0

64+10

______-4/y=0

13

74/13=4/y

74y=13×4

74y= 52

y= 52/74

y= 0.7

Now put valu of y in either quation :

let's take eq1.

13+3/x+2/y=0

13+3/x+2/0.7=0

13+3/x+2.85=0

15.85+3/x=0

15.85= 3/x

15.85x= 3

x= 3/15.85

x= 0.18

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