-10x—16=14 -16x -6 = 14

Mathematics

1 answer:

Alla [95]2 years ago

4 0

Answer: No solution

Step-by-step explanation:

-10x-16=14

-16x-6=14

-10x-16=-16x-6

-10x+16x-16=-16x+16x-6

6x-16=-6

6x-16+16=-6+16

6x=10

x=10/6

x=5/3

-16(5/3)-6=

-80/3-6*3/3=

-80/3-18/3=

(-80-18)/3=-98/3 $\neq$ 14

Hence, there is no solution

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a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

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Step-by-step explanation:

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Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

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$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$