Solve for x & y: x^2+y^2=20. 3x+y=2

2 answers:

6 0

You're solving a system here, easy peasy. Solve the second equation for y to get y = 2 - 3x. Now in the first equation, sub in this y value where you see a y and now the equation is in terms of x only: x^2 + (2-3x)^2 = 20. You have to FOIL out the (2-3x)^2 first to get x^2 + 4 - 12x + 9x^2 = 20. Now combine like terms to get 10x^2 - 12x + 4 = 20. Move the 20 over by subtraction to get
10x^2 - 12x - 16 = 0. Factor this now, taking a 2 out first: 2(5x^2 - 6x -8)=0 and you get factors of 2(x-2)(5x+4)=0. x=2 or x = -4/5. If x = 2, then y = -4. If x = -4/5, then y = 22/5

Lerok [7]4 years ago

5 0

Answer:

Step-by-step explanation:

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