Short answer: C Let's start with the pen. The length is x The width is 3 more.= x + 3 Perimeter of the pen = 2L + 2W P = 2 L + 2W P = 2*x + 2(x + 3) P = 2x + 2x + 6 P = 4x + 6. Leave that for now.
Area of the pen without the doghouse = x(x + 3) Area of the pen without the doghouse = x^2 + 3x Area with the doghouse x^2 + 3x - y = 178 (that's his running around space.
Go to the dog house for a Moment The perimeter of the doghouse and the area are equal and we are told to make them both y. Area = y Perimeter = y
Combined facts. The perimeter of the Pen is three times larger than that of the dog house. 4x + 6 = 3y
The area of the Pen - The area of the doghouse = 178 x^2 + 3x - y = 178
Now what happens is that you have to juggle these two equations so they look like your answer x^2 + 3x - y = 178 4x + 6 = 3y Subtract 3y from both sides. 4x - 3y + 6 = 0 Subtract 6 from both sides. 4x - 3y = - 6
Of 5 out of 12 (5/12) are sports games and 3 out of 12 (3/12) are educational, then you would add these together because you need a total for both because it would include one or the other, so both.
