Respuesta :
We have a rectangular room with the following dimensions:
[tex]\begin{gathered} \text{Length = L} \\ \text{Side = w} \end{gathered}[/tex]The dimensions of the rectangular room is related to each other by the following statement:
" 3 meters longer than it is wide "
We will go ahead an decrypt the above statement and express it into a mathematical form in terms of the dimensions ( defined above ) of the rectangular room:
[tex]L\text{ = w + 3 }\ldots\text{ }\text{\textcolor{#FF7968}{Eq 1}}[/tex]The next statement relates the perimeter ( P ) of the rectangular room in terms of its previously defined dimensions.
" perimeter is 30 meters "
We will go ahead an decrypt the above statement and express it into a mathematical form in terms of the dimensions ( defined above ) of the rectangular room:
[tex]P\text{ = 30 }[/tex]The perimeter of a rectangle ( P ) is defined as the sum of all the boundary sides which can be expressed in terms of its length ( L ) and width ( w ) as follows:
[tex]\begin{gathered} P\text{= 2}\cdot(L+w)\text{ } \\ P=\text{ 2}\cdot(L\text{ + w ) = 30} \\ L\text{ + w = }\frac{30}{2} \\ \\ L\text{ + w = 15 }\ldots\text{\textcolor{#FF7968}{ Eq 2}} \end{gathered}[/tex]Now we have two equations ( Eq1 and Eq2 ) which are just mathematical decryption of the statements provided in the question and these equations are expressed in terms of two variables ( L and w ).
We have two equations and two unknowns we can solve these equations simultaneously.
Step 1: Substitute Eq1 into Eq2.
[tex]\begin{gathered} L\text{ = w + 3 }\ldots\text{\textcolor{#FF7968}{ Eq1}} \\ L\text{ + w = 15 }\ldots\text{ }\text{\textcolor{#FF7968}{Eq2}} \\ =========== \\ (\text{ w + 3 ) + w = 15 }\ldots\textcolor{#FF7968}{Eq3} \end{gathered}[/tex]Step 2: Solve the Eq3 for the variable ( w ).
[tex]\begin{gathered} 2\cdot w\text{ = 15 - 3} \\ w\text{ = }\frac{12}{2} \\ \textcolor{#FF7968}{w}\text{\textcolor{#FF7968}{ = 6 meters}} \end{gathered}[/tex]Step 3: Back-substitution of the result of one variable ( w ) into Eq1.
[tex]\begin{gathered} L\text{ = w + 3 , w = 6 } \\ L\text{ = 6 + 3} \\ \textcolor{#FF7968}{L}\text{\textcolor{#FF7968}{ = 9 meters}} \end{gathered}[/tex]The solution to the two equations are given as:
[tex]\begin{gathered} \textcolor{#FF7968}{L}\text{\textcolor{#FF7968}{ = 9 meters}} \\ \textcolor{#FF7968}{w}\text{\textcolor{#FF7968}{ = 6 meters}} \end{gathered}[/tex]Answer:
