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lilavasa [31]
3 years ago
15

What are the values of x, y, and z in this system of equations?

Mathematics
1 answer:
scZoUnD [109]3 years ago
5 0

Answer:

{x,y,z}={5,−3,3}

Step-by-step explanation:

  [1]    2x + 4y + z = 1

  [2]    x - 2y - 3z = 2

  [3]    x + y - z = -1

// Solve equation [3] for the variable  y  

 [3]    y = -x + z - 1

// Plug this in for variable  y  in equation [1]

  [1]    2x + 4•(-x +z -1) + z = 1

  [1]    -2x + 5z = 5

// Plug this in for variable  y  in equation [2]

  [2]    x - 2•(-x +z -1) - 3z = 2

  [2]    3x - 5z = 0

// Solve equation [2] for the variable  x  

 [2]    3x = 5z

 [2]    x = 5z/3

// Plug this in for variable  x  in equation [1]

  [1]    -2•(5z/3) + 5z = 5

  [1]    5z/3 = 5

  [1]    5z = 15

// Solve equation [1] for the variable  z  

  [1]    5z = 15

  [1]    z = 3

// By now we know this much :

   x = 5z/3

   y = -x+z-1

   z = 3

// Use the  z  value to solve for  x  

   x = (5/3)(3) = 5

// Use the  x  and  z  values to solve for  y  

 y = -(5)+(3)-1 = -3

Solution :

{x,y,z} = {5,-3,3}

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Answer:

C

Step-by-step explanation:

We want to integrate:

\displaystyle \int\frac{4x^4+3}{4x^5+15x+2}\,dx

Notice that the expression in the denominator is quite similar to the expression in the numerator. So, we can try performing u-substitution. Let u be the function in the denominator. So:

u=4x^5+15x+2

By differentiating both sides with respect to x:

\displaystyle \frac{du}{dx}=20x^4+15

We can "multiply" both sides by dx:

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And divide both sides by 5:

\displaystyle \frac{1}{5}\, du=4x^4+3\,dx

Rewriting our original integral yields:

\displaystyle \int \frac{1}{4x^5+15x+2}(4x^4+3\, dx)

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\displaystyle =\int \frac{1}{u}\Big(\frac{1}{5} \, du\Big)

Simplify:

\displaystyle =\frac{1}{5}\int \frac{1}{u}\, du

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\displaystyle =\frac{1}{5}\ln|u|

Back-substitute. Of course, we need the constant of integration:

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Our answer is C.

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