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I believe the general way of solving these kinds of equations is to substitute the values of the coordinate pair for x and y in the equation, and see if the two sides match up. For example, if one of the given coordinate pairs is (2,5) Then, we will substitute x = 2, and y = 5. So, the equation will become: 2^2 + 5^2 = 157*2 + 5 = 17 29 = 319 = 17 This is obviously not true, so (2,5) is not the solution
is it (3, 10)
Try substituting x = 3 and y = 10 in your equation to find out!
First of all you have two = signs in here... is that correct?
Oh is it two equations. x^2 + y^2 = 157 AND x + y = 17
Then the answer (x,y) will be where the x + y = 17 and the x^2 + y^2 = 17
sorry x^2 + y^2 = 157
is it (3,10)
does 3 + 10 = 17
no 30 Idk , i took a wild Guess of tha answer
Oh.. I thought you had multiple choice.
If you have to work it out... it is a bit of a long process... here we go.
x + y = 17 y = 17 - x we substitute that into the other equation. x^2 + y^2 = 157 x^2 + (17-x)^2 = 157 x^2 + (17-x)(17-x) = 157 you foil the 17-x x^2 + 289 - 17x - 17x + x^2 = 157 2x^2 - 34x + 289 = 157 (add like terms) 2x^2 - 34x + 132 = 0 (make it = to 0) 2(x^2 - 17x + 66) = 0 take out gcf 2(x - 11)(x - 6) = 0 factor x - 11 = 0 or x - 6 = 0 x = 11 or x = 6 answers are (11,6) or (6,11)
Actually is is (11,6) AND (6,11) you are intersecting a circle with a line.. That would be two points.