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You have

  1. y = 4x + 8
  2. x² + 7x - 2 = 0
The second is a quadratic involving only x so solve {2} first to find the x values, which can be substituted into the first to find the corresponding y value.

x² + 7x - 2 = 0

→ x = (-7 ± √(7² - 4×1×-2))/(2×1) = (-7 ± √57)/2

→ x = (-7 - √57)/2

→ y = 4 × (-7 - √57)/2 + 8 = -6 - 2√57

or x = (-7 + √57)/2

→ y = 4 × (-7 + √57)/2 + 8 = -6 + 2√57

The solution is the two ordered pairs (x, y)::

( (-7 - √57)/2, -6 - 2√57) and ((-7 + √57)/2, -6 + 2√57)

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6y ago
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6y ago

They are [-7 - sqrt(57)/2, -6 - 2*sqrt(57)] and [-7 + sqrt(57)/2, -6 + 2*sqrt(57)]2.

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Q: What is the solution to the system of equations y equals 4x plus 8 and x squared plus 7x - 2 equals 0?
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