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Is this the correct way to carry out the operation?

Is this the correct way to carry out the operation?-example-1

1 Answer

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Part A: From the diagram of the room, it is clear that the width is w=x+4.

Let l be the length of the room so it follows:


\begin{gathered} l=2x+6+x-4 \\ l=(2x+x)+(6-4) \\ l=3x+2 \end{gathered}

Hence the expression for the total area of the room if the area is A is given by:


\begin{gathered} A=lw \\ A=(3x+2)(x+4) \\ A=3x^2+2x+12x+8 \\ A=3x^2+14x+8 \end{gathered}

Part B: It is given that the length is 12 units so l=12, therefore it follows:


\begin{gathered} 12=3x+2 \\ 12-2=3x-2 \\ 10=3x \\ x=(10)/(3) \end{gathered}

So the area from part A at x=10/3 will be:


\begin{gathered} A=3x^2+14x+8 \\ A=3((10)/(3))^2+14((10)/(3))+8 \\ A=3*(100)/(9)+(140)/(3)+8 \\ A=(300)/(9)+(140*3)/(3*3)+(8*9)/(1*9) \\ A=(300+420+72)/(9) \\ A=(792)/(9) \\ A=88\text{ sq ft} \end{gathered}

Hence the total area of both rooms is 88 square feet.

User Seslyn
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