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2 years ago

14.What is the solution of the system?

Mathematics

1 answer:

Archy [21]2 years ago

3 0

Answer:

14. x=-2.5, y = -7

15. x=28 y = -20

Step-by-step explanation:

14. Let's solve this system by elimination. Multiply the first equation by -1.

-1*(4x-y)= -1*(- 3)

-4x +y = 3

Then add this to the second equation.

-4x+y = 3

6x-y= - 8

--------------

2x = -5

Divide each side by 2

x = -2.5

We still need to find y

-4x+y =3

-4(-2.5) + y =3

10 +y =3

Subtract 10 from each side.

y = 3-10

y = -7

15.I will again use elimination to solve this system, because using substitution will give me fractions which are harder to work with. I will elimiate the y variable. Multiply the first equation by 11

11( 5x+6y)= 11*20

55x+66y = 220

Multiply the second equation by -6

-6(9x+11y)=32*(-6)

-54x-66y = -192

Add the modified equations together.

55x+66y = 220

-54x-66y = -192

---------------------------

x = 28

We still need to solve for y

5x+6y = 20

5*28 + 6y =20

140 + 6y = 20

Subtract 140 from each side

6y = -120

Divide by 6

y = -20

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Let $e^c = C$

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Now from the question we are told that

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From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

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Now taking the exponent of both sides

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=> $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

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divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

$x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

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