General Idea:
(i) Assign variable for the unknown that we need to find
(ii) Sketch a diagram to help us visualize the problem
(iii) Write the mathematical equation representing the description given.
(iv) Solve the equation by substitution method. Solving means finding the values of the variables which will make both the equation TRUE
Applying the concept:
Given: x represents the length of the pen and y represents the area of the doghouse
Statement 1: "The pen is 3 feet wider than it is long"
[tex] Length \; of\; the \; pen = x\\ Width \; of\; the\; pen=x+3 [/tex]
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Statement 2: "He also built a doghouse to put in the pen which has a perimeter that is equal to the area of its base"
[tex] Area \; of\; the\; Dog \; house=y\\ Perimeter \; of\; Dog\; house=y [/tex]
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Statement 3: "After putting the doghouse in the pen, he calculates that the dog will have 178 square feet of space to run around inside the pen."
[tex] Area \; of \; the\; Pen - Area \;of \;the\;Dog \;House=\;Space\;inside\;Pen\\ \\ x \cdot (x+3)-y=178\\ Distributing \;x\;in\;the\;left\;side\;of\;the\;equation\\ \\ x^2+3x-y=178\Rightarrow\; 1^{st}\; Equation\\ [/tex]
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Statement 4: "The perimeter of the pen is 3 times greater than the perimeter of the doghouse."
[tex] Perimeter\; of\; the\; Pen=3\; \cdot \; Perimeter\; of\; the\; Dog\; House\\ \\ 2(x \; + \; x+3)=3 \cdot y\\ Combine\; like\; terms\; inside\; the\; parenthesis\\ \\ 2(2x+3)=3y\\ Distribute\; 2\; in\; the\; left\; side\; of\; the\; equation\\ \\ 4x+6=3y\\ Subtract \; 6\; and \; 3y\; on\; both\; sides\; of\; the\; equation\\ \\ 4x+6-3y-6=3y-3y-6\\ Combine\; like\; terms\\ \\ 4x-3y=-6 \Rightarrow \; \; 2^{nd}\; Equation\\ [/tex]
Conclusion:
The systems of equations that can be used to determine the length and width of the pen and the area of the doghouse is given in Option B.
[tex] 178=x^2+3x-y\\ \\ -6=4x-3y [/tex]
