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Vincent Schneider
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∙ 14y agoYou can't solve a single equation with 2 unknowns.
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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)
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)
Factor 18x3y5 plus 9xy3-3xy?
18x3y5 + 9xy3 - 3xy = 3xy(6x2y4 + 3y2 - 1)
If X equals -1 and y equals 2 what does 2x3-3xy equal?
x = -1 y = 2 2x3 - 3xy = 2 (-1)3 - 3(-1)(2) = 2 (-1) - (-6) = -2 + 6 = 4 Suggestion: Please be careful around problems like this until you have some more experience.
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 of 2xy plus y-3xy?
-1
What are the solutions to the simultaneous equations of x -2y equals 1 and 3xy -y2 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 or 5y^2 +3y -8 = 0 Factorizing the above: (y-1)(5y+8) = 0 meaning y = 1 or y = -8/5 Solutions by substitution are: (3, 1) and (-11/5, -8/5)
What are the points of intersection of the line x -2y equals 1 with the curve 3xy -y squared equals 8 showing work?
If: x -2y = 1 then x = 1+2y If: 3xy -y^2 = 8 then 3(1+2y)y -y^2 = 8 So: 3y+6y^2 -y^2 = 8 => 3y+5y^2 -8 = 0 Solving the above quadratic equation: y = 1 or y = -8/5 By substitution the points of intersection are at: (3, 1) and (-11/5, -8/5)
What is the factor of 3x-6xy?
The GCF is 3xy.
How do you solve the simultaneous equations of x -2y equals 1 and 3xy -y squared equals 8?
x - 2y = 1 → x = 1 + 2y 3xy - y² = 8 Substitute the first equation into the second equation and subtract 8 from both sides: 3(1 +2y)y - y² = 0 3y +6y² - y² - 8 = 0 5y² + 3y - 8 = 0 When factored: (5y +8)(y -1) = 0 Therefore: y = -⁸/₅ or y = 1 Substituting the above values into the linear equation gives the solutions as: (3, 1) and (-¹¹/₈, -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
3x y - 2y?
3xy-2y=0 3xy=2y y=2y (3x) y/2y=3x 1/2=3x multiply across by 2 1=6x 1/6=x therefore substituting x=1/6 into 3xy-2y; 3(1/6)y-2y=0 1/2y=2y y=2y/0.5 0.5 aka 1/2 y=1
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)
