# How to find x and y in simultaneous equations? y^2-4xy+4y=1 3x^2-2xy=1

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### 2 Answers

We have to solve the simultaneous equation

y^2 - 4xy + 4y = 1 ...(1)

3x^2 - 2xy = 1 ...(2)

We see that it is not possible to isolate x in terms of y or y in terms of x using any of the equations.

In (1) we get y = 1/(y - 4x + 4)

In (2) we get x = 1/(3x - 2y)

As this is the case, we cannot the solve the system of simultaneous equations. Using a random solution of (1 , 1), we do get both

y^2 - 4xy + 4y = 1 and 3x^2 - 2xy = 1

**The required solution of the equations is (1, 1)**

We'll change the first equation into:

y^2 + 4y = 4xy + 1

We'll change the 2nd equation into:

3x^2 = 2xy + 1

We'll re-write the first equation as:

y^2 + 4y = 2xy + 1 + 2xy

We'll substitute the sum 2xy + 1 by 3x^2:

y^2 + 4y = 3x^2 + 2xy

We notice that we'll substitute x and y by 1, we'll get:

1^2 + 4*1 = 3*1^2 + 2*1*1

1 + 4 = 3 + 2

5 = 5 true

**The solution of the simultaneous equations is (1 ; 1), where x = 1 and y = 1.**