I uploaded the answer to a file hosting. Here's link:
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Answer:
Step-by-step explanation:
Applying the Laplace transform:
![\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%2B5%5Cmathcal%7BL%7D%5By%27%5D%2B4%5Cmathcal%7BL%7D%5By%27%5D%3D0)
With the formulas:
![\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%3Ds%5E2%5Cmathcal%7BL%7D%5By%5D-y%280%29s-y%27%280%29)
![\mathcal{L}[y']=s\mathcal{L}[y]-y(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%5D%3Ds%5Cmathcal%7BL%7D%5By%5D-y%280%29)
![\mathcal{L}[x]=L](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bx%5D%3DL)
Solving for 
Apply the inverse Laplace transform with this formula:
![\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs-a%7D%5D%3De%5E%7Bat%7D)
![y=3\mathcal{L}^{-1}[\frac1{s+4}]=3e^{-4t}](https://tex.z-dn.net/?f=y%3D3%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs%2B4%7D%5D%3D3e%5E%7B-4t%7D)
1) to this case you must match both equations
3x+2 = 2x-3 ⇒3x-2x = -3-2⇒x= -5
now the value of x substitute it in the any of two equations, y = 3(-5)+2 = -13
and the other equation is the same value
y = 2(-5)-3 = -13
the point is ( -5,-13)
This answer is pretty simple. you see (5,-8) and (20,y). well the 20 is 4 times the 5 on the x value so multiply the y value by 4 to get your answer which Y=-32 in the 2nd corrdenants.
Answer:
Step-by-step explanation:
