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Setler79 [48]
3 years ago
9

System w/Elimination-2x-9y=-25-4x-9y=-25

Mathematics
1 answer:
Lerok [7]3 years ago
5 0

x=0 and y=\frac{25}{9}

Step-by-step explanation:

We need to solve the system by elimination and find values of x and y

-2x-9y=-25\,\,\,(1)\\-4x-9y=-25\,\,\,(2)

Subtract eq(1) and eq(2)

-2x-9y=-25\,\,\,(1)\\-4x-9y=-25\,\,\,(2)\\+\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\---------\\2x=0\\x=0

Now Multiply equation (1) by 2 and subtract with eq(2)

-4x-18y=-50\\-4x-9y=-25\\+\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\--------\\-9y=-25\\y=\frac{-25}{-9}\\y=\frac{25}{9}

So, x=0 and y=\frac{25}{9}

Keywords: system by elimination

Learn more about system by elimination at:

  • brainly.com/question/6075514
  • brainly.com/question/2115716
  • brainly.com/question/4192226

#learnwithBrainly

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Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then 0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow  \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form e^{2t} with t real.

(B) Proceeding and the previous item, we obtain 1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow  \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2. Which is not a vector space with the usual operations (this is because -1/2), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set y=e^{mt} \Rightarrow y''=m^2e^{mt} and we obtain the characteristic equation 0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2 and then the general solution is y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}, and as in the first items the set of solutions form a vector space.

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