Question

solve for x and Y simultaneously 3y + x = 2 & y^2 + x = xy + y

Answer 1

Step-by-step explanation:

$3y + x = 2..........(1) \\ \\ {y }^{2} + x = xy + y \\ {y }^{2} + x = (x + 1)y \\ {y }^{2} = (x + 1)y - x \\ \frac{{y }^{2}}{y} = x + 1 - x \\ y = + 1 \\ \\ substitute \: y = + 1 \: in \: {eq}^{n} (1) \\ \\ 3(1) + x = 2 \\ 3 + x = 2 \\ x = - 1 \\ \\ \\ therefore \: x = - 1 \: and \: y = + 1$

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Answer 2

Question:

Solve for x and y simultaneously:

3y + x = 3

y² + x = xy + y

Answer:

(x = 1/2 , y = 1/2) and (x = -1 , y = 1)

Solution:

We have ;

3y + x = 2 --------(1)

y² + x = xy + y ----------(2)

Using eq-(1) , we have ;

x = 2 - 3y -----(3)

Now,

Putting x = 2 - 3y in eq-(2) , we get ;

=> y² + x = xy + y

=> y² + 2 - 3y = (2 - 3y)•y + y

=> y² + 2 - 3y = 2y - 3y² + y

=> y² + 3y² - 3y - 2y - y + 2 = 0

=> 4y² - 6y + 2 = 0

=> 2•(2y² - 3y + 1) = 0

=> 2y² - 3x + 1 = 0

=> 2y² - 2y - y + 1 = 0

=> 2y(y - 1) - (y - 1) = 0

=> (y - 1)(2y - 1) = 0

=> y = 1 , 1/2

Now,

If y = 1 , then by using eq-(3) , we have ;

=> x = 2 - 3y

=> x = 2 - 3•1

=> x = 2 - 3

=> x = - 1

Again,

If y = 1/2 , then by using eq-(3) , we have ;

=> x = 2 - 3y

=> x = 2 - 3•(1/2)

=> x = 2 - 3/2

=> x = (4 - 3)/2

=> x = 1/2

Hence,

The required solutions are :

(x = 1/2 , y = 1/2) and (x = -1 , y = 1)