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Bingel [31]
3 years ago
9

Solve the following system of equations using Substitution: Submit your answer as an ordered pair. 2x−3y=32x-3y=32x−3y=3 x=y+2x=

y+2x=y+2
Mathematics
1 answer:
exis [7]3 years ago
4 0

9514 1404 393

Answer:

  (3, 1)

Step-by-step explanation:

We assume you want the solution to the system ...

  • 2x−3y=3
  • x=y+2

The second equation gives a nice expression for x, so we can use that in the first equation.

  2(y+2) -3y = 3 . . . . substitute for x in the first equation

  2y +4 -3y = 3 . . . . . eliminate parentheses

  -y = -1 . . . . . . . . . . . collect terms, subtract 4

  y = 1 . . . . . . . . . . . . multiply by -1

  x = 1 +2 = 3 . . . . . . substitute for y in the second equation

The solution is (x, y) = (3, 1).

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<h2>Explanation:</h2>

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Simplest polynomial function is x^4-x^3-11x^2+9x+18.

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Given data:

Zeroes are 3i, –1, 2.

3i is a complex root of the function.

If 3i is a zero of the polynomial then –3i is also a zero of the polynomial.

Therefore zeroes are 3i, –3i –1, 2.

By factor theorem,

If a is zero of the function, then (x – a) is a factor of the polynomial.

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On multiplying the factors, we get the polynomial.

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