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If: 2x+y = 5 then y = 2x-5

If: x^2 -y^2 = 3 then x^2 -(2x-5)^2 = 3

Multiply out the bracket and bringing terms to LHS: -3x^2 -28+20x = 0

Solving the above quadratic equation: x = 14/3 or x = 2

By substitution points of intersection are: (2, 1) and (14/3, -13/3)

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8y ago
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8y ago

2x + y = 5 => y = 5 - 2xThen

x^2 - y^2 = 3 => x^2 - y^2 - 3 = 0

=> x^2 - (5 - 2x)^2 - 3 = 0

=> x^2 - 25 + 20x - 4x^2 - 3 = 0

=> -3x^2 + 20x - 28 = 0

or 3x^2 - 20x + 28 = 0

=> (x - 2)*(3x - 14) = 0

=> x = 2 or x = 4 2/3

x = 2 => y = 1

x = 4 2/3 => y = -4 1/3

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Q: What are the points of intersection of the line 2x plus y equals 5 that crosses the curve x squared -y squared equals 3 showing work?
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