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kap26 [50]
2 years ago
6

–3х - 7y = 8 3х - 2y = 44

Mathematics
2 answers:
lara31 [8.8K]2 years ago
6 0

Answer:

  • One solution

Kind Note:

This question is incomplete. You need to provide the question in order to get the best answer on brainly. In this answer, I will tell you how many solutions are in this system of equations.

Step-by-step explanation:

This question involves the use of 'system of equations'.

  • => -3x - 7y = 8

             3x - 2y = 44

  • => (-3x + 3x) + (-7y - 2y) = (8 + 44)
  • => -9y = 52
  • => y = -52/9 (No further check required)

Since the y-intercept is valid, the system of equations has one solution.

Hence, this system of equations has one solution.

Hoped this helped!

julsineya [31]2 years ago
4 0

Let's solve graphically.

  • -3x-7y=8
  • 3x-2y=44

Graph attached

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Find an n^th degree polynomial with real coefficients satisfying the given conditions. n = 3; -2 and 2 i are zeros; f(-1) = 15.
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well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too

so, the roots are -2, 2i, -2i

now... \bf \begin{cases}&#10;x=-2\implies x+2=0\implies &(x+2)=0\\&#10;x=2i\implies x-2i=0\implies &(x-2i)=0\\&#10;x=-2i\implies x+2i=0\implies &(x+2i)=0&#10;\end{cases}&#10;\\\\\\&#10;(x+2)\underline{(x-2i)(x+2i)}=0\\\\&#10;-----------------------------\\\\&#10;\textit{difference of squares}&#10;\\ \quad \\&#10;(a-b)(a+b) = a^2-b^2\qquad \qquad &#10;a^2-b^2 = (a-b)(a+b)\\\\&#10;-----------------------------\\\\&#10;(x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0&#10;\\\\\\&#10;(x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0&#10;\\\\\\&#10;x^3+2x^2+4x+8=0

now, if we check f(-1), we end up with 5, not 15
hmmm

so, how to turn our 5 to 15? well, 3*5, thus

\bf 3(x^3+2x^2+4x+8)=f(x)\implies 3(5)=f(-1)\implies 15=f(-1)

usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15
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3 years ago
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