Answer:
[tex]on \: this \: farm \: there \: are \to \\ \underline{ \boxed{17 \: pigs }}\: \\ and \\ \underline{ \boxed{28 \:chickens}}[/tex]
Step-by-step explanation:
[tex] \\ \underline{ \boxed{solution \: to \: question \: \boxed{ A}}} \\ \\ let \: the \: pigs \: be \to \: p \\ let \: the \: chicks \: be \to \: c \\let \: the \: total \: animals \: be \to \: 45 \\ hence \to \\ p + c = 45.......eq(1) \\ since \: chickens \: have \: two \: legs : \\ while \: pigs \: have \: four \: legs : \\ let \: animals \: with \: 2 \: legs \: be \to \: 2c \\ let \: animals \: with \: 4\: legs \: be \to \: 4p \\ the \: total \: number \: of \: animal \: legs = 124 \\ hence \to \\ 4p + 2c = 124..........eq(2) \\ therefore \: the \: system \: of \: equations \: \\ to \: describe \: the \: situation \: is \to \\ \\ p + c = 45 \\ 4p + 2c = 124 \\ \\ \\ \underline{ \boxed{solution \: to \: question \: \boxed{B }}} \\ \\ \\ we \: could \: apply \: any \: of \: the \: simultanious \\ principles \to \\ here : am \: appling \: the \: elimination \: method \to \\ p + c = 45 \\ 4p + 2c = 124 \\ to \: elimnate \: one \: varaible \: we \: multiply \: by \to \\ (4) \times p + c = 45 \\ (1) \times 4p + 2c = 124 \\ hence \to \\ - \frac{4p + 4c = 180}{4p + 2c = 124} \\next \to \\ 2c = 56 \\ c = \frac{56}{2} \\ \boxed{ c = 28} \\ lets \: work \: for \: p \to \\ p + c = 45 \\ p + 28 = 45 \\ p = 45 - 28 \\ \boxed{p = 17}[/tex]
