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3 years ago

Simultaneous linear equations

1 answer:

8 0

1) 3x + y = 23
6x + y = 44

Subtract 1st from 2nd,
3x = 21
x = 7
Then, y = 23 - 3(7) = 2

In short, Your Answer would be: (7, 2)

2) 4x - 3y = 21
2x + 3y = 33

Add the equations,
6x = 54
x = 9
Then, 2(9) + 3y = 33
3y = 33 - 18
y = 15/3
y = 5

In short, Your Answer would be: (9, 5)

3) 5x + 2y = 13
6y + 5x = 29
Subtract 1st from 2nd equation,

4y = 16
y = 4
Then, 5x = 13 - 2(4)
x = 5/5 = 1

In short, Your Answer would be: (1, 4)

4) 5x - 4y = 2
10x - 4y = 32
Subtract 1st from 2nd equation,
5x = 30
x = 6

Then, 5(6) - 2 = 4y
y = 28/4 = 7

In short, Your Answer would be: (6, 7)

5) 2x + 7y = 48
x + 3y = 21

Here, x = 21-3y
Substitute in in 1st equation,
2(21-3y) + 7y = 48
42 - 6y + 7y = 48
y = 6

Then, x = 21 - 3(6) = 3

In short, Your Answer would be: (3, 6)

6) 8x - 3y = 44
2y - 4x = -20

Multiply 2nd equation by 2
-8x + 4y = -40
Now, add it to 1st equation,
y = 4

Then, 2(4) - 4x = -20
4x = 8 + 20
x = 28/4 = 7

In short, Your Answer would be: (7, 4)

Hope this helps!

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For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

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Answer:

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Step-by-step explanation:

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