Question

Solve the following systems by Elmination by Substitution method
2y+5/x=3 ; 4y+3/x= 23/5 , x is not = 0
Answer x= 5 , y = 1

Answer 1

Elimination method:

2y + 5/x = 6 ------ (1)

4y + 3/x = 23/5  ---- (2)

On solving (1)*2 & (2), we get

4y + 10/x = 6

4y + 3/x = 23/5

------------------

        7/x = 7/5

         x = 5.



Substitute x =5 in (1), we get

2y + 5/x = 3

2y + 5/5 = 3

2y + 1 = 3

2y = 3 - 1

2y = 2

y = 1.



Therefore x = 5 and y = 1.



Substitution method:

2y + 5/x = 3 

2y = 3 - 5/x

2y = 3x - 5/x

y = 3x - 5/2x ------- (1)


Substitite (1) in (2), we get

4(3x - 5/2x) + 3/x = 23/5

LCM =  5x

10(3x - 5) + 15 = 23x

30x - 55 + 15 = 23x

30x - 30 = 23x

7x = 35

x = 5.


Substitute x = 5 in (1), we get

y = 3x - 5/2x

y = 3(5) - 5/2(5)

y = 15 - 5/10

y = 10/10

y = 1.


Therefore x = 5 and y = 1.


Hope this helps!