Exercises: systems with two variables

Transforming the problem setting into equations

You may denote the number of chickens by some variable $x$ and the number of rabbits by a variable $y$. What is known? A chicken has two legs, a rabbit has four legs. Thus $x$ chickens have $2\cdot x$ legs and $y$ rabbits have $4\cdot y$ legs. This statement gives you equation $\mathrm{I}$:

"There are three times more chickens than rabbits" must also be cast into an equation. Since $y$ is the number of rabbits, you must multiply $y$ by $3$ to get the number $x$ of chickens. So equation $\mathrm{II}$ reads:

You have now obtained the following system of linear equations with two equations and two variables:

To solve this system of equations, you have three methods at your disposal:

• Substitution method

• Elimination by equating coefficients

• Addition method

Here the substitution method is suitable, since the equation $\mathrm{II}$ is already solved for $x$ .

Solution by substitution

Plug equation $\mathrm{II}x=3\cdot y$ into $\mathrm{I}$ :

$\begin{array}{rcll}2\cdot (3y)+4y& =& 120& |\text{ multiply out bracket}\\ & & & \\ 6y+4y& =& 120& |\text{ conclude}\\ & & & \\ 10y& =& 120& |:10\\ & & & \\ y& =& 12& \end{array}$

Plug $y=12$ into equation $\mathrm{II}$ : $x=3\cdot 12=36$

So the solution set of your system of equations is : $𝕃=\{(36|12)\}$

Answer: The farmer has $36$ chickens and $12$ rabbits.

When solving systems of equations, a verification is useful at the end of the calculation. In this case: $36$ chickens have $36\cdot 2=72$ legs and $12$ rabbits have $12\cdot 4=48$ legs. The sum of the legs is $120$, as indicated in the problem setting. The number of chickens was supposed to be three times the number of rabbits. This statement is also correct, since $36=3\cdot 12$. So the given solution set is indeed a solution of the linear system.