Question 703427

There are {{{38}}} animals in a farmyard. Some are {{{cows=x}}} and some are {{{chickens=y}}}. In total there are {{{104}}} legs.

cows have four legs, so it will be {{{4x}}} cows' legs 

chickens have two legs, so it will be {{{2y}}} chickens' legs 

since there are {{{104}}} legs, we have

{{{4x+2y=104}}}......eq. 1

since given that there are {{{38}}} animals in a farmyard, we have

{{{x+y=38}}}.........eq. 2

now solve this system:

{{{4x+2y=104}}}......eq. 1
{{{x+y=38}}}.........eq. 2
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start with {{{x+y=38}}}.........eq. 2 and solve for {{{x}}}

{{{x=38-y}}}.....substitute it in eq. 1

{{{4(38-y)+2y=104}}}......eq. 1...solve for {{{y}}}

{{{152-4y+2y=104}}}

{{{152-2y=104}}}

{{{152-104=2y}}}

{{{48=2y}}}

{{{48/2=y}}}

{{{highlight(24=y)}}}....so, there are {{{24}}} chickens

go back to {{{x=38-y}}} and find {{{x}}}

{{{x=38-24}}}

{{{highlight(x=14)}}}