Please solve and EXPLAIN how to reach the final answer!

If the sum of the sides of a right triangle is 49 inches and the hypoteneuse is 41 inches, find the two sides.

Let x and y denote the two legs. Since the sum of the legs is 49, we have:
x + y = 49. . . . . . . . . . . . . . .(1)

Then, since the hypotenuse is 41, we have that:
x^2 + y^2 = 41^2, by the Pythagorean Theorem
==> x^2 + y^2 = 1681. . . . . . .(2)

By solving (1) for y, we have:
y = 49 - x.

Then, substituting this into (2) gives:
x^2 + (49 - x)^2 = 1681
==> x^2 + (x^2 - 98x + 2401) = 1681
==> 2x^2 - 98x + 2401 = 1681
==> 2x^2 - 98x + 720 = 0
==> 2(x - 40)(x - 9), by factoring
==> x = 9 and x = 40.

When x = 9 and x = 40, y = 40 and y = 9, respectively. Regardless, the two legs are 9 and 40.

I hope this helps!

40in and 9in
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We have a system of equations
a + b = 49
and by the Pythagorean Theorem;
a² + b² = 41²

We can solve b by b = 49 - a

a² + (49 - a)² = 1681
a² + 2401 - 98a + a² = 1681
2a² - 98a + 720 = 0
Divide all terms by2...
a² - 49a + 360 = 0

Solve for quadratics equation however you like... here factoring...

Finding 2 numbers add up to (-49a) and multiply to (a²)(360)
(a - 40)(a - 9) = 0
a = 40 or 9

Therefore, the 2 sides are 40in and 9in.