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3 years ago

Solve using elimination. 5x + 4y = 8 -5x 3y = -1 Submit

Mathematics

1 answer:

ycow [4]3 years ago

6 0

5x + 4y = 8

-5x + 3y = -1

Let's add both equations, to eliminate x.

5x + 4y - 5x + 3y = 8 - 1

7y = 7

y = 1

Let us substitute this value of y in equation 1, so we can find x.

5x + 4 = 8

5x = 4

x = 4/5

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larisa86 [58]

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3 years ago

Consider the initial value problem yâ€²+2y=4t,y(0)=8.

Xelga [282]

Answer:

Please read the complete procedure below:

Step-by-step explanation:

You have the following initial value problem:

$y'+2y=4t\\\\y(0)=8$

a) The algebraic equation obtain by using the Laplace transform is:

$L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\$

next, you replace (1) and (2):

$sY(s)-y(0)+2Y(s)=\frac{4}{s^2}\\\\sY(s)+2Y(s)-8=\frac{4}{s^2}$ (this is the algebraic equation)

$sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)}$ (this is the solution for Y(s))

$y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}$

To find the inverse Laplace transform of the first term you use partial fractions:

$\frac{4}{s^2(s+2)}=\frac{-s+2}{s^2}+\frac{1}{s+2}\\\\=(\frac{-1}{s}+\frac{2}{s^2})+\frac{1}{s+2}$

Thus, you have:

$y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}$

(this is the solution to the differential equation)

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3 years ago

URGENT

IgorLugansk [536]

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4 years ago

4x+6=2x+10 find value of x makes equation true

forsale [732]

Answer:

x = 2

Step-by-step explanation:

$4x + 6 = 2x + 10 \\ \\ 4x - 2x = 10 - 6 \\ \\ 2x = 4 \\ \\ x = \frac{4}{2} \\ \\ x = 2$