0
More values are required to solve this question since there are only two unknowns given. To solve this equation, there has to be only a single unknown value. In this example, assuming x is 0,
0 = y2 + 2y
0 = y (y+2)
y = 0 or y+2 = 0
y = 0 pr y = -2
Therefore, in this case, the points of intersection are (0,0) and (0,-2)
If x=y
Y = y2 + 2y
0 = y2 + y
0 = y(y+1)
y = 0 or y+1 = 0
y = 0 or y=-1
When y=0
x=0
When y=-1
x=-1
Therefore, in this case, the points of intersection are (0,0) and (-1,-1)
Note: to solve quadratic equations, cubic equations or any equation with any unknowns not to the power of 1, you must get a value of 0 on one side of the equation before you can get the final values.
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What are the points of intersection of y equals x squared plus 2x plus 2 with y equals x cubed plus 2?
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
What are the point of intersection of x is equal to y squared plus 2y?
You need two, or more, curves for points of intersection.
What is the point of intersection of the parabolas of y equals 3x squared plus 10x plus 11 and y equals 2 -2x -x squared?
They intersect at the point of: (-3/2, 11/4)
What are the points of intersection of x plus y equals 7 with x squared plus y squared equals 25 showing key stages of work?
If: x+y = 7 and x2+y2 = 25 Then: x = 7-y and so (7-y)2+y2 = 25 => 2y2-14y+24 = 0 Solving the quadratic equation: y = 4 and y = 3 By substitution points of intersection: (3, 4) and (4, 3)
What are the coordinates for x2 plus y2 equals 185 and x plus y equals 17?
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
What are points of intersection of the line 3x -y equals 5 with the curve of 2x squared plus y squared equals 129?
Straight line: 3x-y = 5 Curved parabola: 2x^2 +y^2 = 129 Points of intersection works out as: (52/11, 101/11) and (-2, -11)
What are the points of intersection of y equals x squared plus 2x plus 2 with y equals x cubed plus 2?
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
What are the point of intersection of x is equal to y squared plus 2y?
You need two, or more, curves for points of intersection.
What are the points of intersection of the parabolas of y equals 4x squared -12x -3 and y equals x squared plus 11x plus 5 on the Cartesian plane?
The points are (-1/3, 5/3) and (8, 3).Another Answer:-The x coordinates work out as -1/3 and 8Substituting the x values into the equations the points are at (-1/3, 13/9) and (8, 157)
What is the point of intersection of the parabolas of y equals 3x squared plus 10x plus 11 and y equals 2 -2x -x squared?
They intersect at the point of: (-3/2, 11/4)
What are the points of intersection of x plus y equals 7 with x squared plus y squared equals 25 showing key stages of work?
If: x+y = 7 and x2+y2 = 25 Then: x = 7-y and so (7-y)2+y2 = 25 => 2y2-14y+24 = 0 Solving the quadratic equation: y = 4 and y = 3 By substitution points of intersection: (3, 4) and (4, 3)
Where are the points of intersection on the curve and line of x squared plus y squared plus 4x plus 6y minus 40 equals 0 and x equals 10 plus y?
x2+y2+4x+6y-40 = 0 and x = 10+y Substitute the second equation into the first equation: (10+y)2+y2+4(10+y)+6y-40 = 0 2y2+30y+100 = 0 Divide all terms by 2: y2+15y+50 = 0 (y+10)(y+5) = 0 => y = -10 or y = -5 Substitute the above values into the second equation to find the points of intersection: Points of intersection are: (0, -10) and (5, -5)
What are the coordinates for x2 plus y2 equals 185 and x plus y equals 17?
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
What are the points of intersection of the line 2x plus y equals 5 with x squared minus y squared equals 3 on the Cartesian plane showing work?
Improved Answer:-If: 2x+y = 5 and x^2 -y^2 = 3Then by rearranging: y = 5 -2x and -3x^2 -28+20x = 0Solving the above quadratic equation: x = 2 and x = 14/3By substitution points of intersection are: (2, 1) and (14/3, -13/3)
1 squared plus 1 squared plus 2 squared plus 3 squared plus 5 squared plus 8 squared equals?
104
What are the points of intersection of the line 3x -y equals 5 with 2x squared plus y squared equals 129 showing work?
If: 3x-y = 5 and 2x2+y2 = 129 Then: 3x-y = 5 => y = 3x-5 And so: 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0 Using the quadratic equation formula: x = 52/11 and x = -2 By substitution points of intersection are: (52/11, 101/11) and (-2, -11)
What are the points of intersection of the parabolas y equals 4x squared -2x -1 and -2x squared plus 3x plus 5 showing key stages of work?
If: y = 4x2-2x-1 and y = -2x2+3x+5 Then: 4x2-2x-1 = -2x2+3x+5 So: 6x2-5x-6 = 0 Solving the quadratic equation: x = -2/3 or x = 3/2 Points of intersection by substitution: (-2/3, 19/9) and (3/2, 5)
