2x + y = 5 . . . thus y = 5 - 2x
x2 - y2= 3 . . . . now substitute for y in this equation.
x2 - (5 - 2x)2 = 3 . . . and after expansion this becomes,
x2 - 25 + 20x - 4x2 = 3
-3x2 + 20x - 28 = 0 . . . and this factors as,
(-3x + 14)(x - 2) = 0
Then, x = 2 , x = 14/3
When x = 2, then 2x + y = 5 gives the result y = 1.
When x = 14/3 then 2x + y = 5 gives the result y = -13/3.
If: n squared -n -90 = 0 Then the solutions are: n = 10 or n = -9
If: x-2y = 1 and 3xy -y^2 = 8 Then: x = 1+2y and 3(1+2y)y -y^2 = 8 Hence: 3y+6y^2 -y^2 = 8 => 5y^2 +3y -8 = 0 Solving the above quadratic equation: y =1 or y = -8/5 Solutions by substitution are: when y=1 then x=3 and when y=-8/5 then x=-11/5
5
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)
9 squared is another way of stating the equation 9 X 9, and 10 squared is another way of stating the equation 10 X 10. Find the answers to these equations, add them together, and you will have the answer.
The quadratic equation will have two solutions.
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)
No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2
The solutions are: x = 4, y = 2 and x = -4, y = -2
One of its terms will be squared and it will have two solutions.
No. It's a quadratic equation, and it has two solutions.
The solutions to the quadratic equation are: x = -1 and x = 6
There's an equals missing...
If: 2x+y = 5 and x2-y2 = 3 Then the solutions work out as: (2, 1) and ( 14/3, -13/3)
2x2 - 6x - 25 = 0. Solutions are 5.34 and -2.34
If: x = 2y-2 and x^2 = y^2+7 Then: (2y-2)^2 = y^2+7 => 4y^2-8y+4 = y^2+7 => 3y^2-8y-3= 0 Solving the above quadratic equation: y = -1/3 or y = 3 Solutions by substitution: when y=-1/3 then x=-8/3 and when y=3 then x=4
They are: (3, 1) and (-11/5, -8/5)