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Simora [160]
2 years ago
12

Solve the system of equations below. y=x-1 x+y=3

Mathematics
1 answer:
Airida [17]2 years ago
4 0

Answer:

x=2, y=1

Step-by-step explanation:

y=x-1

y=3-x

x-1=3-x

2x=4

x=2

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For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.

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We have to:x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}

Substituting we have:

x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (5)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-40}} {4}\\x = \frac {4 \pm \sqrt {-24}} {4}

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x = \frac {4 \pm \sqrt {24i ^ 2}} {4}\\x = \frac {4 \pm i \sqrt {24}} {4}\\x = \frac {4 \pm i \sqrt {2 ^ 2 * 6}} {4}\\x = \frac {4 \pm 2i \sqrt {6}} {4}\\x = \frac {2 \pm i \sqrt {6}} {2}

Thus, we have two roots:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

Answer:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

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