MATH HELP! how do you answers this question

The hypotenuse of a right triangle has length 17 cm. The sum of the legs is 23 cm. What are their lengths?
I don't know how to solve this question...please help!

if you call one leg = x then other leg =(23-x)
x^2 +(23-x)^2=17^2
x^2+x^2-46x+23^2=289
2x^2-46x+529=289
2x^2-46x+240=0
x^2-23x+120=0 (x-15)(x-8)=0
giving the legs as 15cm and8cm

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Call the legs x and y.

From the Pythagorean Theorem (PT)

x^2 + y^2 = 17^2

Also,

x + y = 23
so
y = 23 - x


Substitute in the PT

x^2 + (23 - x)^ = 17^2

x^2 + x^2 - 46x + 529 = 289

2x^2 - 46x + 240 = 0

x^2 - 23x + 120 = 0

(x-8)(x-15) = 0

x 8 or 15


replace in x+y = 23 and solve for y

y = 15 or 8

So the lengths of the legs are 8cm and 15cm.

15cm & 8cm
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Let the 2 legs be x and y

x + y = 23
so y = 23 - x

According to the Pythagorean Theorem:
x² + (23 - x)² = 17²

Open brackets with FOIL
x² + (529 - 46x + x²) = 289
2x² - 46x + 240 = 0 <===Standard Quadratic Equation

By completing the square

x² - 23x + 120 = 0
x² - 23 = -120
x² - 23 + (-23)/2 ² = -120 + (-23)/2 ²
x² - 23 + 529 / 4 = - 480 / 4 + 529 / 4
(x - 23/2)² = 49 / 4
√(x - 23/2)² = ±√49 / √4
x - 23/2 = ±7 / 2
x = (23±7) / 2

1) x = (23 + 7) / 2 = 30 / 2 = 15, y = 23 - 15 = 8
2) x = (23 - 7) / 2 = 16 / 2 = 8, y = 23 - 8 = 15

The lengths of the legs of the right triangle are 15cm and 8cm.

Just write down equations for given info.
x^2 + y^2 = 17^2 Pythagoras
x + y = 23
Now use y = 23 - x to substitute
x^2 + (23^2 - 46x + x^2) = 17^2
tidy that up a bit
x^2 - 23x + 120 = 0
(x - 8)(x - 15) = 0
So x can be either 8 or 15
then y can be either 15 or 8