Question

solve for 'x' and' y' , if x²-y²=3(x+y)=5(x-y)​

Answer 1

3(x+y)=5(x-y)

(x+y)/5=(x-y)/3

3x+3y=5x-5y

3x-5x=-3y-5y

-2x=-8y

x=4y

x^2-y^2=3(x+y)...(given)

=(x+y)(x-y)=3(x+y)...(x^2-y^2=(x+y)(x-y))

substitute x=4y

4y+y×3y=12y+3y

15y^2=15y15y^2-15y=0

on solving thisy=0or y=1

substitute y=0 we get x=0 and y=1 we get x= 4

therefore x=0,y=0 and x=4andy=1 is the answer

hope you like it

Answer 2

Answer:

(1) : x² - y² = 3

(2) : xy = 2 → y = 2/x

Restart from the equation (1)

x² - y² = 3 → you know that: y = 2/x

x² - (2/x)² = 3

x² - (4/x²) = 3

(x^4 - 4)/x² = 3

x^4 - 4 = 3x²

x^4 - 3x² - 4 = 0 → let: X = x² → where X ≥ 0

X² - 3X - 4 = 0

X² - (4X - X) - 4 = 0

X² - 4X + X - 4 = 0

(X² - 4X) + (X - 4) = 0

X.(X - 4) + (X - 4) = 0

(X + 1).(X - 4) = 0

First case: (X + 1) = 0 → X + 1 = 0 → X = - 1 ← no possible because the condition

Second case: (X - 4) = 0 → X - 4 = 0 → X = 4 ← possible

Recall: X = x² → x² = 4 → x = ± 2

First solution: x = 2

y = 2/x

→ y = 1

Second solution: x = - 2

y = 2/x

→ y = - 1