This morning, some dragons, horses, and chickens were playing in my backyard. I counted 15 heads, 50 legs, and 4 wings. How many dragons, how many horses, and how many chickens were playing in my backyard? (For your information: any healthy dragon has three heads, four legs, and two wings. Feel free to use Wikipedia to obtain the information about the number of heads, legs, and wings of the rest of the creatures. Oh, and all creatures on my backyard were healthy.)

Accepted Solution

We are given number of heads =15we know that any healthy dragon has three headshorse has 1 headchicken has 1 headLet's assume number of dragons is xnumber of horses is ynumber of chickens is zso, we will get first equation:[tex] 3x+y+z=15 [/tex]number of legs =50any healthy dragon has four legschicken has 2 legshorse has four legsso, we can get second equation as[tex] 4x+4y+2z=50 [/tex]we can simplify it [tex] 2(2x+2y+z)=50 [/tex][tex] 2x+2y+z=25 [/tex]now, we can find third equation dragon has two wings horse has no wings chicken has two wings so, we will get third equations as[tex] 2x+0y+2z=4 [/tex]now, we can simplify it [tex] 2x+2z=4 [/tex][tex] 2(x+z)=4 [/tex][tex] x+z=2 [/tex]so, we will get system of equations as[tex] 3x+y+z=15 [/tex][tex] 2x+2y+z=25 [/tex][tex] x+z=2 [/tex]now, we can use substitution We can find for z from third equation [tex] z=2-x [/tex]we can plug this in first equation [tex] 3x+y+2-x=15 [/tex]now, we can solve for y [tex] 2x+y+2=15 [/tex][tex] y=13-2x [/tex]now, we can plug this z and y into second equation [tex] 2x+2(13-2x)+2-x=25 [/tex]now, we can solve for x[tex] x-4x+26=23 [/tex][tex] -3x=-3 [/tex][tex] x=1 [/tex]now, we can find y and z[tex] y=13-2x [/tex]we can plug x=1[tex] y=13-2*1 [/tex][tex] y=11 [/tex][tex] z=2-x [/tex]we can plug x=1[tex] z=2-1 [/tex][tex] z=1 [/tex]Hence , number of dragons is 1number of horses is 11number of chicken is 1............Answer