Question

What values of (x,y) satisfy the given equation x^3-2^x^2y+2xy^2-y^2=0

Answer 1

Answer:

Equation I = 2x + 3y = 5

Equation II = 7x - 4y = 3

Multiplying first equation by 4 =

→ 2x + 3y = 5

∴ (2x + 3y = 5) x (4)

= (8x + 12y = 20)......... (equation.III)

Now Multiplying second equation by 3 =

→ 7x - 4y = 3

∴ (7x - 4y = 3) x (3)

= (21x - 12y = 9).......... (equation IV)

Now, Add Equation III with Equation IV.

→ (8x + 12y = 20) + (21x - 12y = 9)

→ (8x + 21x) + (12y - 12y) = (20 + 9)

→ 29x + 0 = 29

→ 29x = 29

→ x = 29/29

→ x = 1

We get : x = 1,

Now Putting x = 1 in Equation I,

→ 2x + 3y = 5

→ 2(1) + 3y = 5

→ 2 + 3y = 5

→ 3y = 5 - 2

→ 3y = 3

→ y = 3/3

→ y = 1

So (x, y) = (1, 1)

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Answer 2

Answer:

(x, y) = (1, 1)

Explanation: