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Question Number 168053 by lapache last updated on 01/Apr/22

Solve the system { ((x+2y+3z+2t=2)),((2x+5y−8z+6t=5)),((3x+4y−5z+2t=4)) :}

Solvethesystem{x+2y+3z+2t=22x+5y−8z+6t=53x+4y−5z+2t=4

Answered by MJS_new last updated on 01/Apr/22

3 equations; 4 variables ⇒ parametric solution ((x),(y),(z),(t) ) = (((2p)),((1−2p)),(0),(p) ) with p∈R

3equations;4variables⇒parametricsolution(xyzt)=(2p1−2p0p)withp∈R

Answered by FelipeLz last updated on 03/Apr/22

x+2y+3z+2t+2x+5y−8z+6t−(3x+4y−5z+2t) = 2+5−4 3x+7y−5z+8t−3x−4y+5z−2t= 3 3y+6t = 3 y = 1−2t • 3x+4y−5z+2t = 4 3x+4y−5z+2t = 2(x+2y+3z+2t) 3x+4y−5z+2t = 2x+4y+6z+4t x−11z−2t = 0 z = 0 → x−11(0)−2t = 0 x = 2t • 2x+5y−8z+6t−2(x+2y+3z+2t) = 5−2(2) 2x+5y−8z+6t−2x−4y−6z−4t = 1 y−14z+2t = 1 y = 1−2t → 1−2t−14z+2t = 1 14z = 0 ⇒ z = 0 • x = α → y = 1−α; z = 0; t = (α/2) y = β → x = 1−β; z = 0; t = ((1−β)/2) t = γ → x = 2γ; y = 1−2γ; z = 0 S = {(x, y, z, t)∣ (x, y, z, t) = (𝛂, 1−𝛂, 0, (α/2)) ∀𝛂 ∈ R} S = {(x, y, z, t)∣ (x, y, z, t) = (1−𝛃, 𝛃, 0, ((1−𝛃)/2)) ∀𝛃 ∈ R} S = {(x, y, z, t)∣ (x, y, z, t) = (2𝛄, 1−2𝛄, 0, 𝛄) ∀𝛄 ∈ R}

x+2y+3z+2t+2x+5y−8z+6t−(3x+4y−5z+2t)=2+5−43x+7y−5z+8t−3x−4y+5z−2t=33y+6t=3y=1−2t∙3x+4y−5z+2t=43x+4y−5z+2t=2(x+2y+3z+2t)3x+4y−5z+2t=2x+4y+6z+4tx−11z−2t=0z=0→x−11(0)−2t=0x=2t∙2x+5y−8z+6t−2(x+2y+3z+2t)=5−2(2)2x+5y−8z+6t−2x−4y−6z−4t=1y−14z+2t=1y=1−2t→1−2t−14z+2t=114z=0⇒z=0∙x=α→y=1−α;z=0;t=α2y=β→x=1−β;z=0;t=1−β2t=γ→x=2γ;y=1−2γ;z=0S={(x,y,z,t)∣(x,y,z,t)=(α,1−α,0,α2)∀α∈R}S={(x,y,z,t)∣(x,y,z,t)=(1−β,β,0,1−β2)∀β∈R}S={(x,y,z,t)∣(x,y,z,t)=(2γ,1−2γ,0,γ)∀γ∈R}

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