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Sagot :
To solve the exercise you can propose the following system of linear equations:
[tex]\begin{gathered} x+y=23\Rightarrow\text{ Equation 1} \\ 32x+26y=688\text{ }\Rightarrow\text{ Equation 2} \\ \text{ Where} \\ x\text{ is the number of double rooms and} \\ y\text{ is the number of single rooms} \end{gathered}[/tex]To solve the system of linear equations you can use the method of reduction or elimination. To do this first multiply Equation 1 by -32 and add both equations, then solve for the variable y:
[tex]\begin{gathered} (x+y)\cdot-32=23\cdot-32 \\ -32x-32y=-736 \\ \text{ Now add both equations} \\ -32x-32y=-736 \\ 32x+26y=688\text{ +} \\ ------------- \\ 0x-6y=-48 \\ -6y=-48 \\ \text{ Divide by -6 from both sides of the equation} \\ \frac{-6y}{-6}=\frac{-48}{-6} \\ y=8 \end{gathered}[/tex]Finally, replace the value of the variable y into any of the initial equations and solve for the variable x. For example, replace the value of the variable y in Equation 1:
[tex]\begin{gathered} x+y=23 \\ x+8=23 \\ \text{ Subtract 8 from both sides of the equation} \\ x+8-8=23-8 \\ x=15 \end{gathered}[/tex]Therefore, 15 double rooms and 8 single rooms were rented.
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