First of all let's write collaboration time formula:
"t_C=\\dfrac{V}{\\upsilon_1+\\upsilon_2}" , where tC - collaboration time, V - volume of the work, "\\upsilon_1" - labor productivity of the first man, "\\upsilon_2" - labor productivity of the second man.
Now let's write labor productivity formulas for both men:
"\\upsilon_1=\\dfrac{V}{t_1}" and "\\upsilon_2=\\dfrac{V}{t_2}" , where V - volume of the work, t1 - first man-hours used, t2 - second man-hours used.
Let's put these formulas in collaboration time formula:
"t_C=\\dfrac{V}{\\dfrac{V}{t_1}+\\dfrac{V}{t_2}}=\\dfrac{V}{V(\\dfrac{1}{t_1}+\\dfrac{1}{t_2})}="
"=\\dfrac{1}{\\dfrac{1}{t_1}+\\dfrac{1}{t_2}}=\\dfrac{1}{\\dfrac{t_2+t_1}{t_1*t_2}}=\\dfrac{t_1*t_2}{t_1+t_2}"
Let's mark t1 as x, then t2 will be x - 1. Collaboration time is equal to 3 hours. Now we can write equation:
"\\dfrac{x*(x-1)}{x+(x-1)}=3"
Now let's simplify and solve it:
"\\dfrac{x^2-x}{2x-1}=3"
"x^2-x=3(2x-1)"
"x^2-x=6x-3"
"x^2-7x+3=0"
a = 1, b = -7, c = 3
"x_1=\\dfrac{-b-\\sqrt{b^2-4ac}}{2a}="
="\\dfrac{-(-7)-\\sqrt{(-7)^2-4*1*3}}{2*1}=\\dfrac{7-\\sqrt{49-12}}{2}="
"=\\dfrac{7-\\sqrt{37}}{2}\\approx0.5" hours does not satisfy the conditions of our task because man alone can't clean house faster than both men together (0.5 < 3)
"x_2=\\dfrac{-b+\\sqrt{b^2-4ac}}{2a}="
="\\dfrac{-(-7)+\\sqrt{(-7)^2-4*1*3}}{2*1}=\\dfrac{7+\\sqrt{49-12}}{2}="
"=\\dfrac{7+\\sqrt{37}}{2}\\approx6.5" hours is man-hours of the first guy.
6.5 - 1 = 5.5 hours is man-hours of the second guy.
Answer: 6.5 hours and 5.5 hours or 6 hours 30 mins and 5 hours 30 mins.
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