Exercises: systems with two variables

A hotel has 105 beds, which are located in 40 two- and three-bed rooms. How many two-and-three-bed rooms does the hotel have?

Solve the problem by using a system of equations!

For this task you need the following basic knowledge: System of linear equations

Extracting equations from the problem setting

What is known? The sum of the number of two-bed rooms $x$ and the number of three-bed rooms $y$ is 40. This statement gives you equation $I$ :

I x + y = 40_{}

In a double room there are two beds. Thus, there are $2 \cdot x$ beds in $x$ two-bed rooms and correspondingly, there are $3 \cdot x$ beds in the three-bed rooms. In total, the hotel has 105 beds, so you can set up equation $I I$ :

II 2 x + 3 y = 105

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

\begin{array}{cl} I x & + & y & = & 40 \\ II 2 x & + & 3 y & = & 105 \end{array}

Solve the system

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

Solving by substitution method

Solve equation $II$ for one of the two variables and substitute this variable into equation $II$ . It doesn't matter whether you solve $II$ for $x$ or $y$ .

$I^{'} y = 40 - x$ . plugged into $II :$

\begin{array}{rcl} 2 \cdot x + 3 \cdot (40 - x) & = & 105 & | multiply out bracket \\ 2 x + 120 - 3 x & = & 105 & | conclude \\ - x + 120 & = & 105 & | - 120 \\ - x & = & - 15 & | \cdot (- 1) \\ x & = & 15 \end{array}

Plug $x = 15$ into equation $I^{'}$ : $y = 40 - 15 = 25$

So the solution set of the linear system is : $𝕃 = {(15 | 25)}$

Answer: The hotel has 15 two-bed rooms and 25 theree-bed rooms.

Solving by equating coefficients

Solve both equations for the same variable. You have already solved the equation $I$ for $y$ in the substitution method:

$I^{'} y = 40 - x$ . Now, also $II$ must be solved for $y$ :

\begin{array}{rcl} 2 \cdot x + 3 \cdot y & = & 105 & | - 2 x \\ 3 y & = & 105 - 2 x & | : 3 \\ I I^{'} y & = & 35 - \frac{2}{3} x \end{array}

Now set equal the two right sides of the equations $I^{'}$ and $I I^{'}$ :

\begin{array}{rcl} 40 - x & = & 35 - \frac{2}{3} x & | + x \\ 40 & = & 35 + \frac{1}{3} x & | - 35 \\ 5 & = & \frac{1}{3} x & | \cdot 3 \\ x & = & 15 \end{array}

Plug $x = 15$ into $I^{'}$ : $y = 40 - 15 = 25$

So the solution set of your system of equations is : $𝕃 = {(15 | 25)}$

Answer: The hotel has 15 two-bed rooms and 25 theree-bed rooms.

Solvin by the addition method

\begin{array}{cccccccl} I x & + & y & = & 40 \\ II 2 x & + & 3 y & = & 105 \end{array}

The goal is that by adding the two equations, one of the unknown variables cancels out. In this system of equations you can, for instance, multiply equation $I$ by $(- 2)$ , which gets you $I^{'}$ . By adding $I^{'}$ and $II$ , you get the solution for $y$ :

\begin{array}{cl} I^{'} & - 2 x & - & 2 y & = & - 80 \\ II & 2 x & + & 3 y & = & 105 \\ I^{'} + II & 0 x & + & y & = & 25 \end{array}

Plug $y = 25$ into $I$ : $x + 25 = 40 \Rightarrow x = 15$

So the solution set of the linear system is : $𝕃 = {(15 | 25)}$

Answer: The hotel has 15 two-bed rooms and 25 theree-bed rooms.

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