Right triangle with legs x and x - 7, and hypotenuse x + 1.

We are given a right triangle with sides:

- One leg = $ x $
- Other leg = $ x - 7 $
- Hypotenuse = $ x + 1 $

Since it’s a right triangle, we can use the Pythagorean Theorem:

> $ \text{leg}_1^2 + \text{leg}_2^2 = \text{hypotenuse}^2 $

So plug in the expressions:

$ x^2 + (x - 7)^2 = (x + 1)^2 $

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Now expand each term.

First, $ (x - 7)^2 = x^2 - 14x + 49 $

Second, $ (x + 1)^2 = x^2 + 2x + 1 $

So the equation becomes:

$ x^2 + (x^2 - 14x + 49) = x^2 + 2x + 1 $

Simplify left side:

$ x^2 + x^2 - 14x + 49 = 2x^2 - 14x + 49 $

Right side stays: $ x^2 + 2x + 1 $

Now set them equal:

$ 2x^2 - 14x + 49 = x^2 + 2x + 1 $

Subtract $ x^2 + 2x + 1 $ from both sides to get everything on one side:

$ (2x^2 - 14x + 49) - (x^2 + 2x + 1) = 0 $

Distribute the minus sign:

$ 2x^2 - 14x + 49 - x^2 - 2x - 1 = 0 $

Combine like terms:

- $ 2x^2 - x^2 = x^2 $
- $ -14x - 2x = -16x $
- $ 49 - 1 = 48 $

So:

$ x^2 - 16x + 48 = 0 $

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Now solve this quadratic equation.

Factor it:

We need two numbers that multiply to 48 and add to -16.

Try: -12 and -4 → (-12)(-4) = 48, (-12)+(-4) = -16 → YES!

So:

$ (x - 12)(x - 4) = 0 $

Set each factor to zero:

- $ x - 12 = 0 $ → $ x = 12 $
- $ x - 4 = 0 $ → $ x = 4 $

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Now check which solution makes sense in context.

Remember the side lengths must be positive.

Check $ x = 4 $:

- Leg 1: $ x = 4 $
- Leg 2: $ x - 7 = 4 - 7 = -3 $ → Not possible! Side length can’t be negative.

So discard $ x = 4 $.

Check $ x = 12 $:

- Leg 1: $ 12 $
- Leg 2: $ 12 - 7 = 5 $
- Hypotenuse: $ 12 + 1 = 13 $

Check Pythagorean theorem: $ 12^2 + 5^2 = 144 + 25 = 169 = 13^2 $ → Correct!

Also, all sides positive → valid.

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Final Answer:
12