There is a right triangle. The hypotenuse is 17 units. The sum of the other two sides is 23. Find the length of the two other sides.

Thanks for everyone's help in advance!

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$a+b=23 \implies b=23-a$.

Note that $a^{2}+b^{2}=17^{2}$ and that $a+b=23$.

Substituting $b=23-a$ in $a^{2}+b^{2}=17^{2}$ gets you:

$a^{2}+(a-23)^{2}=289 \implies 2a^{2}-46a+240=0$

$a=\frac{46 \pm \sqrt {46^{2}-(4)(2)(240)}}{(2)(2)} =$ $\frac{46 \pm \sqrt {46^{2}-(1920)}}{(4)}\implies$$a=15 or a=8. Then you plug the a value into the first equation and you get that one of the sides is 15 units and the other side is 8 units. - Well, we know that$$a+b=23 \qquad\text{and}\qquad a^2+b^2=17^2=289.$$Substituting b=23-a into the second equation will give you a quadratic equation in a. It's quite likely that of the two solutions to that quadratic equation, only one will be positive. There's your value of a, and then b=23-a recovers b. - From the Pythagorean theorem, you have a^2+b^2=c^2 with c=17, a+b=23. This is three equations in three unknowns. - There are only a few sets of small integers that form right triangles. These are:$$(3,4,5)\\ (5,12,13)\\ (7,24,25)\\ (8,15,17)$$and multiples of these, for example$2\cdot(3,4,5) = (6,8,10)$. You should memorize these, because teachers like to use them over and over again. In this case the problem mentions that the hypotenuse is 17, so you should immediately ask if this is the$(8,15,17)$triangle. Since$8+15=23\$, it is and you have the answer and can go on to the next problem faster than the person at the next desk.