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wlad13 [49]
3 years ago
6

Solve the following quadratic-linear system of equations.

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
3 0

For this case we have a system of two equations with two unknowns.


To solve, we equate both equations:


2x-2 = x ^ 2-x-6\\x ^ 2-3x-4 = 0

We have a quadratic equation of the form:


x ^ 2 + bx + c = 0

Where:


a = 1\\b = -3\\c = -4

Its roots are given by:


x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}\\x = \frac {- (- 3) \pm \sqrt {(- 3) ^ 2-4 (1) (- 4)}} {2 (1)}\\x = \frac {3 \pm \sqrt {9 + 16}} {2}\\x = \frac {3 \pm \sqrt {25}} {2}\\x = \frac {3 \pm5} {2}\\

We have two roots:


x_ {1} = \frac {3 + 5} {2} = 4\\x_ {2} = \frac {3-5} {2} = - 1

Thus, we look for the values of y, by substituting any of the equations:


For\ x_ {1} = 4,\ y_ {1} = 2 (4) -2 = 6\\For\ x_ {2} = - 1,\ y_ {2} = 2 (-1) -2 = -4

Answer:


(x_ {1}, y_ {1}) = (4,6)\\(x_ {2}, y_ {2}) (- 1, -4)

Option B

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3 years ago
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Vlad [161]
<h2>Answer: 13 & 19</h2>

<h3>Step-by-step explanation:</h3>

Let the bigger number be 'U' and the smaller number 't'.

Since the two numbers have a sum of 32, then   U + t = 32.

& since one number is 6 more than the other, the U - 6 = t

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By substituting fot the value of U we found,

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Step-by-step explanation:

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Answer:

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