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13
(9,-2)
-- Pick any number at all and call it " x " . Any number whatsoever. -- Multiply it by 8 and call the product " y " . -- There you are ! Now you have an ordered pair that's a solution to that equation. There's no limit. You can keep inventing ordered pairs just like that, until you have millions or billions of them, or as many as you want.
This can be done with the equation (x1+x2)/2, (y1+y2)/2 which, when solved, creates a (x,x) solution, or a coordinate pair solution. if you had the points (2,4) and (4,8) you would put x1 (2) plus (+) x2 (4) divided by 2, and 2+4 is 6, and 6/2 is 3, so we know our midpoint x value is 3. Then, we would plug in our 'y' values, so we would have y1 (4) + y2 (8) and 4+8 = 12 and 12/2 is 6, so our solution coordinate ordered pair would be (3,6).
x^2 - 15x + 56 = 0
1/8
y=(-1) x=(2)
Without an equality sign the given terms can't be considered to be a straight line equation.
13
4x-2y=16 has the solution points (0,-8) and (4, 0) at the coordinate axes, and wherever y= 4(x-4).
8
(9,-2)
-- Pick any number at all and call it " x " . Any number whatsoever. -- Multiply it by 8 and call the product " y " . -- There you are ! Now you have an ordered pair that's a solution to that equation. There's no limit. You can keep inventing ordered pairs just like that, until you have millions or billions of them, or as many as you want.
This can be done with the equation (x1+x2)/2, (y1+y2)/2 which, when solved, creates a (x,x) solution, or a coordinate pair solution. if you had the points (2,4) and (4,8) you would put x1 (2) plus (+) x2 (4) divided by 2, and 2+4 is 6, and 6/2 is 3, so we know our midpoint x value is 3. Then, we would plug in our 'y' values, so we would have y1 (4) + y2 (8) and 4+8 = 12 and 12/2 is 6, so our solution coordinate ordered pair would be (3,6).
x^2 - 15x + 56 = 0
I assume you mean (8, 0). If one or both of the coordinates are zero, the point is not in any of the four quadrants. Instead, it is on the axes - between two quadrants.
It is 2.