Answer:
Step-by-step explanation:
Let's use the definition of the Laplace transform and the identity given:![\mathcal{L}[t \cos 5t]=(-1)F'(s)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bt%20%5Ccos%205t%5D%3D%28-1%29F%27%28s%29)
with ![F(s)=\mathcal{L}[\cos 5t]](https://tex.z-dn.net/?f=F%28s%29%3D%5Cmathcal%7BL%7D%5B%5Ccos%205t%5D)
.
Now, 
. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that 
.
Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that
.
Solving for F(s) on the last equation, 
, then the Laplace transform we were searching is 
X+y = 4, therefore y = 4-x when you rearrange the equation.
You can subsitute this in: x-(4-x) = 6
-4+x = 6-x
x=10-x
2x = 10
x = 5
Substitute x into the previous equation:
5+y = 4
y = 4-5
y = -1
34x=0
So we need to isolate x. Since 34 is multiplied by x, we need to isolate x by moving 34 over the equal sign. Whenever you move something over the equal sign, the opposite is done. For 34, instead of being multiplied, it will be divided.
34x=0
X=0/34
Then you solve for x
X=0
Hope this helps!
Answer:
D Multiply equation B by 5 and equation A by 4 and add the results together
