Tractors

Two tractors plow the field in 4 hours together. If the first tractor plow half of the field and then the second tractor completed the job, it would take 9 hours. How many hours does the field plow for each tractor separately?

Result

a =  12 h
b =  6 h

Solution:

4 (1/a+1/b)=1  a/2+b/2=9 a+b=18 b=18a  4 (b+a)=ab=0  4 ((18a)+a)=a(18a) a218a+72=0  p=1;q=18;r=72 D=q24pr=1824172=36 D>0  a1,2=q±D2p=18±362 a1,2=18±62 a1,2=9±3 a1=12 a2=6   Factored form of the equation:  a=(a12)(a6)=0a=a1=12=12  h 4 \cdot \ (1/a+1/b) = 1 \ \\ \ \\ a/2 + b/2 = 9 \ \\ a + b = 18 \ \\ b = 18-a \ \\ \ \\ 4 \cdot \ (b+a) = ab = 0 \ \\ \ \\ 4 \cdot \ ((18-a)+a) = a(18-a) \ \\ a^2 -18a +72 = 0 \ \\ \ \\ p = 1; q = -18; r = 72 \ \\ D = q^2 - 4pr = 18^2 - 4\cdot 1 \cdot 72 = 36 \ \\ D>0 \ \\ \ \\ a_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 18 \pm \sqrt{ 36 } }{ 2 } \ \\ a_{1,2} = \dfrac{ 18 \pm 6 }{ 2 } \ \\ a_{1,2} = 9 \pm 3 \ \\ a_{1} = 12 \ \\ a_{2} = 6 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ a=(a -12) (a -6) = 0a = a_{ 1 } = 12 = 12 \ \text{ h }

Checkout calculation with our calculator of quadratic equations.

b=18a=1812=6=6  h b = 18 -a = 18 -12 = 6 = 6 \ \text{ h }



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