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System Solution Sum (Posted on 2024-11-18)

x and y are real numbers such that x² + 3y = 9, and 2y² - 6x = 18. If the 4 solutions to the system are (x₁,y₁), (x₂,y₂), (x₃,y₃), and (x₄,y₄), then find (y₁ - x₁) + (y₂ - x₂) + (y₃ - x₃) + (y₄ - x₄).

No Solution Yet Submitted by Danish Ahmed Khan

Rating: 5.0000 (1 votes)

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Solution

| Comment 2 of 3 |

Solve the first equation for y: y = (-1/3)x^2 + 3

Substitute this into the second equation and simplify: x^4 - 18x^2 - 27x = 0

The coefficient of x^3 is 0, thus x1+x2+x3+x4=0.

Solve the second equation for x: x = (1/3)y^2 - 3

Substitute this into the first equation and simplify: y^4 - 18y^2 + 27y = 0

The coefficient of y^3 is 0, thus y1+y2+y3+y4=0.

Then (y1-x1)+(y2-x2)+(y3-x3)+(y4-x4) = (y1+y2+y3+y4) - (x1+x2+x3+x4) = 0 - 0 = 0.

Posted by Brian Smith on 2024-11-18 11:25:35

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