Consider the pair of equations: $x^{2}-x y-x=22$ and $y^{2}-x y+y=34$. If $x>y$, then $x-y$ equals

We have two equations:

$x^2 - xy - x = 22$ ... (1)

$y^2 - xy + y = 34$ ... (2)

We need to find $x - y$ when $x > y$.

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We add equation (1) and equation (2):

$(x^2 - xy - x) + (y^2 - xy + y) = 22 + 34$

$x^2 - xy - x + y^2 - xy + y = 56$

Combining like terms:

$x^2 + y^2 - 2xy - x + y = 56$

Key insight: Notice that we can rearrange this as:

$x^2 + y^2 - 2xy - (x - y) = 56$

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The first three terms $x^2 + y^2 - 2xy$ form a perfect square pattern.

Since $(x - y)^2 = x^2 - 2xy + y^2$, our equation becomes:

$(x - y)^2 - (x - y) = 56$

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Let $u = x - y$. Then our equation becomes:

$u^2 - u = 56$

$u^2 - u - 56 = 0$

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We need to factor $u^2 - u - 56 = 0$.

This can be written as: $u(u - 1) = 56$

We need two consecutive integers whose product is 56.

Let's find factors of 56:

$56 = 1 \times 56$ (difference = 55, not consecutive)

$56 = 2 \times 28$ (difference = 26, not consecutive)

$56 = 4 \times 14$ (difference = 10, not consecutive)

$56 = 7 \times 8$ (difference = 1, these are consecutive!)

Since $u$ and $(u-1)$ are consecutive integers with product 56, we have:

$u = 8$ and $u - 1 = 7$

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Since $u = x - y = 8$, we have our answer.

We can check that $8 \times 7 = 56$

Also, since $x > y$, we have $x - y > 0$, so $x - y = 8$ (positive) is correct.

$x - y = 8$