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perform the solution of the mathematical problem, given s: x= 3-5r, y= 6+kr, z=-1+4r l: 2x+3y+z= 12, 2x+5y=14 find the real number of K of so that the lines are parallel

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Answer to a math question perform the solution of the mathematical problem, given s: x= 3-5r, y= 6+kr, z=-1+4r l: 2x+3y+z= 12, 2x+5y=14 find the real number of K of so that the lines are parallel

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Seamus
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Resposta = Vamos resolver este sistema de equações para encontrar o valor de \(k\) tal que as retas sejam paralelas. Dado sistema de equações: 1. \(x = 3 - 5r\) 2. \(y = 6 + kr\) 3. \(z = -1 + 4r\) 4. \(2x + 3y + z = 12\) 5. \(2x + 5y = 14\) Primeiro, vamos expressar a segunda equação em termos de \(x\): \[ y = 6 + kr \] \[ 3y = 18 + 3kr \] \[ y = 6 + kr \] Agora, vamos substituir as expressões por \(x\) e \(y\) na terceira equação: \[ z = -1 + 4r\] E a quarta equação: \[2x + 3y + z = 12\] \[ 2(3 - 5r) + 3(6 + kr) + (-1 + 4r) = 12 \] Resolvendo para \(r\): \[ 6 - 10r + 18 + 3kr - 1 + 4r = 12 \] \[ 23 + (3k - 6)r = 12 \] \[ (3k - 6)r = 12 - 23 \] \[ (3k - 6)r = -11 \] \[ r = \frac{-11}{3k - 6} \] Agora, vamos considerar a quinta equação: \[2x + 5y = 14 \] \[ 2(3 - 5r) + 5(6 + kr) = 14 \] Substitua a expressão por \(r\): \[ 2(3 - 5 \cdot \frac{-11}{3k - 6}) + 5(6 + k \cdot \frac{-11}{3k - 6}) = 14 \] Resolvendo para \(k\): \[ 6 + \frac{110}{3k - 6} + 30 + \frac{5k \cdot (-11)}{3k - 6} = 14 \] \[ 36 + \frac{110 - 55k}{3k - 6} = 14 \] \[ 22 = \frac{55k - 110}{3k - 6} \] \[ 22(3k - 6) = 55k - 110 \] \[ 66 mil - 132 = 55 mil - 110 \] \[ 11k = 22\] \[ k = 2 \] Portanto, o valor de \(k\) para o qual as retas são paralelas é **2**.

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