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Question Number 186099 by akuba last updated on 01/Feb/23

1. Solve for the linear system: x−2y+3z=5 2x+y−4z=0 3x+4y−11z=−5 2. Ghana railways co−operation has 20 trains for it′s operations. It is observed that x trains can accommodate 2 passengers, y trains 3 passengers and z trains 5 passengers. Howeer, the totsl number of passrngers always present at Ghana railways are 64. During market day, 3 of x trains, 2 of y trains and 4 of z trains for a total of 10 trains were used. Determine the values of x, y and z.

$$\mathrm{1}.\:{Solve}\:{for}\:{the}\:{linear}\:{system}: \\ $$$${x}−\mathrm{2}{y}+\mathrm{3}{z}=\mathrm{5} \\ $$$$\mathrm{2}{x}+{y}−\mathrm{4}{z}=\mathrm{0} \\ $$$$\mathrm{3}{x}+\mathrm{4}{y}−\mathrm{11}{z}=−\mathrm{5} \\ $$$$\mathrm{2}.\:{Ghana}\:{railways}\:{co}−{operation}\:{has}\:\mathrm{20}\: \\ $$$${trains}\:{for}\:{it}'{s}\:{operations}.\:{It}\:{is}\:{observed}\:{that} \\ $$$${x}\:{trains}\:{can}\:{accommodate}\:\mathrm{2}\:{passengers}, \\ $$$${y}\:{trains}\:\mathrm{3}\:{passengers}\:{and}\:{z}\:{trains}\:\mathrm{5}\: \\ $$$${passengers}.\:{Howeer},\:{the}\:{totsl}\:{number}\:{of} \\ $$$${passrngers}\:{always}\:{present}\:{at}\:{Ghana}\: \\ $$$${railways}\:{are}\:\mathrm{64}.\:{During}\:{market}\:{day},\:\mathrm{3}\:{of} \\ $$$${x}\:{trains},\:\mathrm{2}\:{of}\:{y}\:{trains}\:{and}\:\mathrm{4}\:{of}\:{z}\:{trains}\:{for}\: \\ $$$${a}\:{total}\:{of}\:\mathrm{10}\:{trains}\:{were}\:{used}.\:{Determine}\: \\ $$$${the}\:{values}\:{of}\:{x},\:{y}\:{and}\:{z}. \\ $$

Answered by Mastermind last updated on 01/Feb/23

Ques. 1 x=2, y=0 and z=1 solution Using Row reduction Method x−2y+3z=5 5y−10z=−10 10y−20z=−20 x−2y+3z=5 ....(1) 5y−10z=−10 ...(2) therefore, from (2), we have y=2z−2 putting y in (1), we get x=z+1 Now, z is a free variable taking z as 1 we have x=2, y=0 and z=1 Mastermind

$$\mathrm{Ques}.\:\mathrm{1} \\ $$$$\mathrm{x}=\mathrm{2},\:\:\mathrm{y}=\mathrm{0}\:\mathrm{and}\:\mathrm{z}=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{solution} \\ $$$$\mathrm{Using}\:\mathrm{Row}\:\mathrm{reduction}\:\mathrm{Method} \\ $$$$\mathrm{x}−\mathrm{2y}+\mathrm{3z}=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\mathrm{5y}−\mathrm{10z}=−\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\mathrm{10y}−\mathrm{20z}=−\mathrm{20} \\ $$$$ \\ $$$$\mathrm{x}−\mathrm{2y}+\mathrm{3z}=\mathrm{5}\:\:\:\:\:\:\:\:....\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{5y}−\mathrm{10z}=−\mathrm{10}\:...\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{therefore},\:\mathrm{from}\:\left(\mathrm{2}\right),\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\mathrm{y}=\mathrm{2z}−\mathrm{2} \\ $$$$\mathrm{putting}\:\mathrm{y}\:\mathrm{in}\:\left(\mathrm{1}\right),\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{x}=\mathrm{z}+\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Now},\:\mathrm{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{free}\:\mathrm{variable} \\ $$$$\mathrm{taking}\:\mathrm{z}\:\mathrm{as}\:\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{x}=\mathrm{2},\:\mathrm{y}=\mathrm{0}\:\mathrm{and}\:\mathrm{z}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$

Commented by akuba last updated on 01/Feb/23

please explain the row reduction method

$${please}\:{explain}\:{the}\:{row}\:{reduction}\:{method} \\ $$

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