Question

(a+b)x+(a−b)y = 4a2b2 and (a-b)x+(a+b)y=0 then value of x is :​

Answer 1

Answer:

The value of x is 8a³ - y.

Step-by-step explanation:

We can solve the given system of equations to determine the value of x:

$(a+b)x + (a-b)y = 4a^2b^2 ...(1)$

$(a-b)x + (a+b)y = 0 ...(2)$

We can resolve this system through the process of elimination. The y term can be taken out of equation (1) and equation (2) by adding them together:

$(a+b)x + (a-b)y + (a-b)x + (a+b)y = 4a^2b^2 + 0$

Simplifying the equation's left side:

$2ax + 2ay = 4a^2b^2$

When 2a is subtracted from the two terms on the left side:

$2a(x + y) = 4a^2b^2$

Divide both sides by 2a, and the result is

$x + y = 2ab^2 ......(3)$

The value of y from equation (3) can now be entered into equation (1) as follows:

$(a+b)x + (a-b)(2ab^2 - x) = 4a^2b^2$

Increasing and reducing the equation:

$ax + bx + 2ab^3 - bx - (a-b)x = 4a^2b^2$

Simplifying further:

$2ab^3 = 4a^2b^2$

Dividing both sides by 2ab²:

b = 2a

Now, substituting the value of b back into equation (3):

x + y = 2ab²

x + y = 2a(2a)²

x + y = 8a³

Since we know that x + y = 2ab² = 2a(2a)² = 8a³, we can conclude that x + y = 8a³.

To find the value of x, we need to subtract y from both sides of the equation:

x = 8a³ - y

Therefore, the value of x is 8a³ - y.

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