Eh this is semi hard. explain your thought process/work when done! !

Hello from MrBillDoesMath!

x =2, y= 3, z = 1

Discussion:

Applying the Pythagorean thereon to the leftmost triangle gives:

y^2 + (sqrt(3))^2 =  (sqrt(12) ) ^2     =>

y^2 + 3                = 12                     => subtract 3 from both sides

y^2                      = 12 - 3 = 9          => take square root of both sides

y =   3

In the full triangle, based on similar triangles, we have

(y + z) /  (sqrt(12)) =  sqrt(12)/y   =>  multiply both sides by sqrt(12)

(y + z) =  (sqrt(12)^2 /y               =>   multiply both sides by y

y(y+z) = 12                                  =>  y = 3. Substitute.

3 (3 + z) = 12                              => divide both sides by 4

(3+z) = 12/3 =4                            => subtract 3 from both sides

z = 4 -3 = 1

Finally, using Pythagoras in the small rightmost triangle gives

x^2 = ((sqrt(3))^2 + z^2             => simplify

x^2 = 3 + z^2                            => but z = 1 from above. Substitute

x^2 = 3 + 1^2 = 4                       => take square roots of both sides

x = sqrt(4) = 2

Conclusion: x =2, y= 3, z = 1

Thank you,

MrB

graph them on a graphing calculator, press 2nd trace. select intersect. hit enter until it says answer. ti 84 works best

step-by-step explanation:

use distributive property on all like terms and change your signs if you have a double negative or if you have positive/negative