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.
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 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)
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)
