If: 3x-2y = 1
Then: 9x^2 = (1+2y)^2 by squaring both sides
If: 3x^2 -2y^2 +5 = 0
Then: 9x^2 = 6y^2 -15 by multiplying all terms by 3
So: 6y^2 -15 = 1+4y +4y^2
Transposing and collecting like terms: 2y^2 -16 -4y = 0
Divide all terms by two: y^2 -8 -2y = 0
Factorizing: (y-4)(y+2) = 0 => y = 4 or y = -2
Therefore solutions by substitution are: (3, 4) and (-1, -2)
squared
true
The solutions to the quadratic equation are: x = -1 and x = 6
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
The basic method is the same as for other types of equations: you need to isolate the variable ("x", or whatever variable you need to solve for). In the case of radical equations, it often helps to square both sides of the equation, to get rid of the radical. You may need to rearrange the equation before squaring. It is important to note that when you do this (square both sides), the new equation may have solutions which are NOT part of the original equation. Such solutions are known as "extraneous" solutions. Here is a simple example (without radicals): x = 5 (has one solution, namely, 5) Squaring both sides: x squared = 25 (has two solutions, namely 5, and -5). To protect against this situation, make sure you check each "solution" of the modified equation against the original equation, and reject the solutions that don't satisfy it.
The solutions are: x = 4, y = 2 and x = -4, y = -2
They are: (3, 1) and (-11/5, -8/5)
If: 2x+y = 5 and x2-y2 = 3 Then the solutions work out as: (2, 1) and ( 14/3, -13/3)
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)
These are two expressions, not equations. Expressions do not have solutions, only equations do. NB equations include the equals sign.
The two rational solutions are (0,0,0) and (1,1,1). There are no other real solutions.
Four.
4
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
If: x = 2y-2 and x^2 = y^2+7 Then: (2y-2)^2 = y^2+7 So: (2y-2)(2y-2) = y^2+7 It follows: 4y^2-y^2-8y+4-7 = 0 => 3y^2-8y-3 = 0 Solving the above quadratic equation: y = -1/3 or y = 3 Solutions: when y = 3 then x = 4 and when y = -1/3 then x = -8/3
If you mean: x2+2x+1 = 0 then it is a quadratiic equations whose solutions are equal because x = -1 and x = -1
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