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Find the dimensions of a rectangle whose perimeter is 42 m and whose area is 104 m^2
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2x+2y = 42 xy=104
simultaneous eqns
x=length , y =width
Let x,y be length and width of rectangle 2x+2y = 42 xy = 104 solve system using substitution 2y = 42-2x y = 21 -x x(21-x) = 104 21x - x^2 = 104 x^2 -21x +104 = 0 (x-13)(x-8)=0 x=8 or x=13 dimensions are 8 and 13
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