Question

(a-b) x + (a+b) y = a^{2} - 2ab - b^{2}

(a+b)(x+y) = a^{2} + b^{2}

Select the right options for the values of x and y. A x = a+b B x = a-b C y = \frac{2ab}{a-b} D y = \frac{-2ab}{a+b}

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Answer

The correct options are:

Option A:

$$ x = a + b $$

Option D:

$$ y = \frac{-2ab}{a+b} $$

To find the values of $x$ and $y$, we start with the equations provided:

[ (a - b)x + (a + b)y = a^2 - 2ab - b^2 \quad \text{(1)} ] [ (a + b)(x + y) = a^2 + b^2 \quad \text{(2)} ]

Step 1: Solve for x

Subtract equation (2) from equation (1):

[ (a - b)x + (a + b)y - [(a + b)x + (a + b)y] = (a^2 - 2ab - b^2) - (a^2 + b^2) ]

Simplify the left-hand side:

[ (a - b)x - (a + b)x = -2ab - 2b^2 ]

Combining the $x$ terms:

[ (a - b - a - b)x = -2ab - 2b^2 ]

This simplifies to:

[ -2bx = -2b(a + b) ]

Divide both sides by $-2b$:

[ x = a + b ]

Option A is confirmed: ( x = a + b )

Step 2: Solve for y using equation (1)

Substitute $x = a + b$ into equation (1):

[ (a - b)(a + b) + (a + b)y = a^2 - 2ab - b^2 ]

Since $(a - b)(a + b) = a^2 - b^2$, the equation becomes:

[ a^2 - b^2 + (a + b)y = a^2 - 2ab - b^2 ]

Cancel $a^2 - b^2$ on both sides:

[ (a + b)y = -2ab ]

Divide both sides by $(a + b)$:

[ y = \frac{-2ab}{a + b} ]

Option D is confirmed: ( y = \frac{-2ab}{a + b} )


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