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Solve the system of equations.

x + y = 15

y = x2 – 5

(–5, 20) and (4, 11)
(3, 4) and (7, 8)
(5, 10) and (–4, 19)
(5, 20) and (4, 11)

Sagot :

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Answer:

  (a)  (–5, 20) and (4, 11)

Step-by-step explanation:

The answer choices can be tried in the equations.

  A: -5+20 = 15, 4+11 = 15; (-5)² -5 = 20, 4² -5 = 11 . . . . correct choice

  B: 3+4 ≠ 15 . . . . fails first equation

  C: 5+10 = 15, -4+19 = 15; 5² -5 ≠ 10 . . . . fails second equation

  D: 5+20 ≠ 15 . . . . fails first equation

_____

When there is one linear and one quadratic equation, it usually works well to solve by substitution. Here, we can use the second equation to substitute for y in the first equation:

  x +(x² -5) = 15

  x² +x = 20 . . . . add 5

  x² +x +0.25 = 20.25 . . . . add 0.25 to complete the square

  (x +0.5)² = (±4.5)² . . . . write as squares

  x = -0.5 ± 4.5 . . . . . . . take the square root

  x = -5 or 4 . . . . matches choice A

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