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What are the solutions to the simultaneous equations of x -2y equals 1 and 3xy -y squared equals 8?
They are: (3, 1) and (-11/5, -8/5)
What are the points of intersection of the line of x -2y equals 1 with the curve of 3xy -y squared equals 8 showing work?
If: x-2y = 1 and 3xy-y2 = 8 Then: x =1+2y and so 3(1+2y)y-y2 = 8 => 3y+5y2-8 = 0 Solving the quadratic equation: y = 1 or y = -8/5 Points of intersection by substitution: (3, 1) and (-11/5, -8/5)
What is the answer to 3xy - 3 2xy?
Assuming the two equations are: 3xy - 3 = 0 ............................................. 1 2xy = 0 ............................................. 2 Then from eq 1 xy = 1 from eq 2 xy = 0 That is not possible. Or, assuming the equation is 3xy - 3 + 2xy = 0 Then 5xy = 3 or y = 3/5x Substitute for any value of x to get the value of y
How do you find the points of intersection of the straight line x -2y equals 1 with the curve 3xy - y squared equals 8?
x-2y = 1 => x = 2y+1 3xy-y2 = 8 Substitute x = 2y+1 into the second equation: 3y(2y+1)-y2 = 8 6y2+3y-y2 = 8 5y2+3y-8 = 0 Solving the above with the quadratic equation formula will give y values of: -8/5 and 1 Substitute these values into the first equation to find the values of x: So the points of intersection are: (3, 1) and (-11/5, -8/5)
What are the points of contact when the line x -2y equals 1 meets the curve of 3xy -y squared equals 8?
If: x -2y = 1 then x = 2y+1 If: 3xy -y^2 = 8 then 3*(2y+1)*y -y^2 = 8 So: 6y^2 +3y -y^2 = 8 Collecting like terms and deducting 8 from both sides: 5y^2 +3y -8 = 0 Factorizing: (5y+8)(y-1) = 0 => y = -8/5 or y =1 Points of contact by substitution: (3, 1) and (-11/5, -8/5)
