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Law Incorporation [45]
3 years ago
11

elimination was used to solve a system of equations. one of the intermediate steps led to the equation 2x= 8. which of the follo

wing systems could have led to this equation?

Mathematics
1 answer:
konstantin123 [22]3 years ago
8 0

9514 1404 393

Answer:

  C

Step-by-step explanation:

The step shown indicates that y was eliminated from the equations. For choices A, B, D, this is done by adding the equations together (the y-coefficients are opposites). The result of doing that gives x-terms of 4x, 0x, and 0x, respectively. These x-terms do not match the one given: 2x.

For choice C, the y-term is eliminated by subtracting twice the second equation from the first. Doing that gives ...

  (4x +2y) -2(x +y) = (14) -2(3)

  4x +2y -2x -2y = 14 -6 . . . . eliminate parentheses

  2x = 8 . . . . . . . . . . . . . . collect terms

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3 years ago
What is the prime factorization of 96? 8 · 12 2 · 2 · 2 · 2 · 3 25 · 3 25 · 32
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3 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
7 0
3 years ago
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