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What are the solutions to the simultaneous equations of x squared -xy - y squared equals -11 and 2x plus y equals 1 solved by completing the square?
Wiki User
∙ 7y ago
If: 2x+y = 1 then y =1-2x and y^2 = (1-2x)^2 which is 1-4x+4x^2
If: x^2 -xy -y^2 = -11 then x^2 -x(1-2x) -(1-4x+4x^2) = -11
Expanding brackets: x^2 -x +2x^2 -1 +4x -4x^2 = -11
Collecting like terms and adding 11 to both sides: -x^2 +3x +10 = 0
Dividing all terms by -1: x^2 -3x -10 = 0
Completing the square: (x-3/2)^2 -9/4 -10 = 0 => (x-3/2)^2 = 49/4
Square root both sides: x-3/2 = -/+ 7/2
Add 3/2 to both sides: x = 3/2 -/+ 7/2
Therefore: x = 5 or x = -2
Solutions by substitution are: (5, -9) and (-2, 5)
Wiki User
∙ 7y ago
Wiki User
∙ 7y agoThey are (-2, 5) and (5, -9).
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What are the solutions to the simultaneous equations of x squared plus y squared equals 20 and 2y minus x equals 0?
The solutions are: x = 4, y = 2 and x = -4, y = -2
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 solutions of the simultaneous equations of x squared -xy -y squared equals -11 and 2x plus y equals 1?
1st equation: x^2 -xy -y squared = -11 2nd equation: 2x+y = 1 Combining the the two equations together gives: -x^2 +3x +10 = 0 Solving the above quadratic equation: x = 5 or x = -2 Solutions by substitution: (5, -9) and (-2, 5)
What are the solutions to -x squared plus 2x plus 4 and x-2?
These are two expressions, not equations. Expressions do not have solutions, only equations do. NB equations include the equals sign.
How many solutions can two unique equations that both have two squared variables?
Four.
What are the solutions to the simultaneous equations of x squared plus y squared equals 20 and 2y minus x equals 0?
The solutions are: x = 4, y = 2 and x = -4, y = -2
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 solutions to the simultaneous equations of 2x plus y equals 5 and x squared -y squared equals 3?
If: 2x+y = 5 and x2-y2 = 3 Then the solutions work out as: (2, 1) and ( 14/3, -13/3)
What are the solutions of the simultaneous equations of x squared -xy -y squared equals -11 and 2x plus y equals 1?
1st equation: x^2 -xy -y squared = -11 2nd equation: 2x+y = 1 Combining the the two equations together gives: -x^2 +3x +10 = 0 Solving the above quadratic equation: x = 5 or x = -2 Solutions by substitution: (5, -9) and (-2, 5)
What are the solutions to -x squared plus 2x plus 4 and x-2?
These are two expressions, not equations. Expressions do not have solutions, only equations do. NB equations include the equals sign.
What are 3 real rational solutions to this system of equations z equals x squared y equals z squared x equals y squared?
The two rational solutions are (0,0,0) and (1,1,1). There are no other real solutions.
How many solutions can two unique equations that both have two squared variables?
Four.
Up to how many solutions can there be in a system of two unique equations that both have two squared variables?
4
What are the solutions to the simultaneous equations of 2x plus y equals 5 and x squared minus y squared equals 3?
Merge the equations together and form a quadratic equation in terms of x:- 3x2-20x+28 = 0 (3x-14)(x-2) = 0 x = 14/3 or x = 2 So when x = 14/3 then y = -13/3 and when x = 2 then y = 1
What is another name for solution in equations using x squared 2x plus 1 0?
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What are the solutions to the simultaneous equations of x squared plus xy plus y squared equals 7 and 2x plus y equals 1?
I suggest you solve the second equation for one of the variables, then replace that in the first equation.Another Answer:-x2+xy+y2 = 7 and 2x+y = 1Merging the two equations together in terms of x will result in a quadratic equation of:3x2-3x-6 = 0Divide all terms by 3:x2-x-2 = 0When factored: (x+1)(x-2) = 0So: x = -1 or x = 2Solutions: when x = -1 then y = 3 and when x = 2 then y = -3
Why do equations of the form x2 equals k have no real solutoins when k0?
If the equation is x2 = k0 Then there are two solutions: + or - 1, since any number to 0 is 1 and 1 or -1 squared = 1.
