# What are solutions of simultaneous equations 1/x=(y-1)/y 3/x=(2y-1)/y

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We have to solve the equations 1/x=(y-1)/y and 3/x=(2y-1)/y

1/x=(y-1)/y

=> x = y/(y - 1)

Substitute x = y/ (y - 1) in 3/x=(2y-1)/y

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

=> 3 /[y/ (y - 1)] = (2y-1)/y

=> 3(y - 1)/y = (2y - 1)/y

=> 3y - 3 = 2y - 1

=> y = 2

x = 2/(2 - 1) = 2

**Therefore we get x = 2 and y = 2**

We'll re-write the given equations by cross multiplying:

y = x(y-1)

We'll remove the brackets:

y = xy - x

We'll add x both sides:

x + y = xy (1)

We'll change the 2nd equation in:

3y = x(2y-1)

3y = 2xy - x

We'll add x both sides:

3y + x = 2xy (2)

We'll substitute (1) in (2):

3y + x = 2(x+y)

3y + x = 2x + 2y

3y - 2y = 2x - x

y = x (3)

We'll substitute (3) in (2)

3x + x = 2x^2

4x = 2x^2

We'll divide by 2 and we'll use the symmetric property:

x^2 - 2x = 0

We'll factorize by x:

x(x-2) = 0

x = 0

x-2 = 0

x = 2

So, because x = y => y = 2

**The solution of the system is {2 ; 2}.**