a)
x lesser than 6 = x more than 2
(6 - x) = (2 + x) | - x
b) Method 1
(6 - 2x) = 2 | + 2x
6 = 2 + 2x | - 2
2x=4
x = 2
Method 2
\(\tt \, Area\,of\,Desk\,1\rightarrow (x)(2+x) \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= (2x+x^2) \,ft^2 \\ Area\,of\,Desk\,2 \rightarrow (x)(6-x) \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= (6x-x^2)\,ft^2\\ This\,should\,be\,true:(2x+x^2)=(6x-x^2)\quad|\,-(6x-x^2)\\2x^2-4x=0\\ Two\,possible\,results\,of\,x.x = 0\,and\,x=2.\\ As\,an\,area\,of\,0\,doesn't\,make\,sense,\,then\,x=2.\)
c) \(x=2,\\ 2(2x+x^2) = 2(2 \times 2 + 2^2) \\ = 2(4+4) \\ = 2(8) \\ = 16\)
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