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?

Correct 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)  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:  (a12)(a6)=0 a=a1=12 h4 \cdot \ (1/a+1/b)=1 \ \\ \ \\ a/2 + b/2=9 \ \\ a + b=18 \ \\ b=18-a \ \\ \ \\ 4 \cdot \ (b+a)=ab=0 \ \\ 4*((18-a)+a)=a(18-a) \ \\ \ \\ 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 -12) (a -6)=0 \ \\ a=a_{1}=12 \ \text{h}

Checkout calculation with our calculator of quadratic equations.

b=18a=1812=6 h



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