Question

brainly.in find the values of x and y satisfying the given equation . x^2+y^2-4x=0 and x+y=4

Answer 1

Step-by-step explanation:

x^2+y^2-4x=0

x^2+y^2=4x

(x+y) ^2-2xy=4x

4^2-2xy=4x {(x+y) =4}

16-2xy=4x

16=4x+2xy

Dividing by 2

8=2x+xy

8=x(2+y)

Now, x+y=4

y=4-x

put the value of y in 8=x(2+y)

8=x(2+4-x)

8=x(6-x)

8=6x-x^2

x^2-6x+8=0

Solving the equation by splitting the middle term

x^2-4x-2x+8=0

x(x-4) -2(x-4) =0

(x-2) (x-4) =0

x=2 or x=4

If x=2 , y=2

If x=4 , y=0

Answer 2

Answer:

Step-by-step explanation:

let the fraction be x / y

then acc. to ques.

x + y + 3= 2y

x = y - 3 ............. (i)

also

x-1/y-1 = y/2y = 1/2

x - 1 =( y-1) / 2

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

x = (y +1) / 2

2x = y + 1

2x - y = 1 ............... (ii)

Subtituting the value of (i) in (ii) we have

2 ( y - 3 ) - y =1

2y - 6 -y = 1

y - 6 = 1

y = 7

x = y - 3

x = 7 - 3

x = 4

Therefore the fraction is x / y = 4 / 7

IF IT IS CORRECT MARK BRAINLIEST ✌️