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2 years ago

Please help solve by elimination I hope its not blurry Have a nice day and Goodnight

Mathematics

2 answers:

Rama09 [41]2 years ago

8 0

Answer:

1. x= 13/5,y=-18/5

2. x=4,y=-7

3.x=8,y=-1

4.x=-3,y=-5

Step-by-step explanation:

There you go and have a good day or night

strojnjashka [21]2 years ago

7 0

Answer:

$\displaystyle \textcolor{black}{4.}\:[-3, -5]$

$\displaystyle \textcolor{black}{3.}\:[8, -1]$

$\displaystyle \textcolor{black}{2.}\:[4, -7]$

$\displaystyle \textcolor{black}{1.}\:[3, -2]$

Step-by-step explanation:

When using the Elimination method, you eradicate one pair of variables so they are set to zero. It does not matter which pair is selected:

$\displaystyle \left \{ {{2x - 3y = 9} \atop {-5x - 3y = 30}} \right.$

{2x - 3y = 9

{⅖[−5x - 3y = 30]

$\displaystyle \left \{ {{2x - 3y = 9} \atop {-2x - 1\frac{1}{5}y = 12}} \right. \\ \\ \frac{-4\frac{1}{5}y}{-4\frac{1}{5}} = \frac{21}{-4\frac{1}{5}} \\ \\ \boxed{y = -5, x = -3}$

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$\displaystyle \left \{ {{x - 2y = 10} \atop {x + 3y = 5}} \right.$

{x - 2y = 10

{⅔[x + 3y = 5]

$\displaystyle \left \{ {{x - 2y = 10} \atop {\frac{2}{3}x + 2y = 3\frac{1}{3}}} \right. \\ \\ \frac{1\frac{2}{3}x}{1\frac{2}{3}} = \frac{13\frac{1}{3}}{1\frac{2}{3}} \\ \\ \boxed{x = 8, y = -1}$

_______________________________________________

$\displaystyle \left \{ {{y = -3x + 5} \atop {y = -8x + 25}} \right.$

{y = −3x + 5

{−⅜[y = −8x + 25]

$\displaystyle \left \{ {{y = -3x + 5} \atop {-\frac{3}{8}y = 3x - 9\frac{3}{8}}} \right. \\ \\ \frac{\frac{5}{8}y}{\frac{5}{8}} = \frac{-4\frac{3}{8}}{\frac{5}{8}} \\ \\ \boxed{y = -7, x = 4}$

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$\displaystyle \left \{ {{y = -x + 1} \atop {y = 4x - 14}} \right.$

{y = −x + 1

{¼[y = 4x - 14]

$\displaystyle \left \{ {{y = -x + 1} \atop {\frac{1}{4}y = x - 3\frac{1}{2}}} \right. \\ \\ \frac{1\frac{1}{4}y}{1\frac{1}{4}} = \frac{-2\frac{1}{2}}{1\frac{1}{4}} \\ \\ \boxed{y = -2, x = 3}$

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I am joyous to assist you at any time.

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3 years ago

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Ganezh [65]

Answer:

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

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

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

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

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

Taking inverse of Laplace transformation

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

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

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

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

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