Without an equality sign the given expression is not an equation and so therefore it has no solution.
No solution. Multiply second equation by 3 and get 12x + 6y = 0 but already stated 12x + 6y = 9...
x+y=0 2x+2y=0 This homogeneous system has infinitely many non-trivial solutions. If you are looking for exactly one non-trivial solution, no such system exists. the system may or may not have non trivial solution. if number of variables equal to number of equations and given matrix is non singular then non trivial solution does not exist
One way is to look at the graphs of these equations. If they intersect, the point of intersection (x, y) is the only solution of the system. In this case we say that the system is consistent. If their graphs do not intersect, then the system has no solution. In this case we say that the system is inconsistent. If the graph of the equations is the same line, the system has infinitely simultaneous solutions. We can use several methods in order to solve the system algebraically. In the case where the equations of the system are dependent (the coefficients of the same variable are multiple of each other), the system has infinite number of solutions solution. For example, 2x + 3y = 6 4y + 6y = 12 These equations are dependent. Since they represent the same line, all points that satisfy either of the equations are solutions of the system. Try to solve this system of equations, 2x + 3y = 6 4x + 6y = 7 If you use addition or subtraction method, and you obtain a peculiar result such that 0 = 5, actually you have shown that the system has no solution (there is no point that satisfying both equations). When you use the substitution method and you obtain a result such that 5 = 5, this result indicates no solution for the system.
If an ordered pair is a solution to a system of linear equations, then algebraically it returns the same values when substituted appropriately into the x and y variables in each equation. For a very basic example: (0,0) satisfies the linear system of equations given by y=x and y=-2x By substituting in x=0 into both equations, the following is obtained: y=(0) and y=-2(0)=0 x=0 returns y=0 for both equations, which satisfies the ordered pair (0,0). This means that if an ordered pair is a solution to a system of equations, the x of that ordered pair returns the same y for all equations in the system. Graphically, this means that all equations in the system intersect at that point. This makes sense because an x value returns the same y value at that ordered pair, meaning all equations would have the same value at the x-coordinate of the ordered pair. The ordered pair specifies an intersection point of the equations.
When the solution is equal to a number, in other words, only one solution happens if you solve for your variables. 0 = 0 means there are many solutions and a number not equal to another number has no solution.
It works out that x = 0 and y = 3
For (x, y) = (0, 3) the first equation gives: 2*0 + 3 = -3 that is, +3 = -3 which is FALSE. So (0, 3) is not a solution.
I) x>0 II) y>0 The first quadrant is the part of the coordinate plane where x and y are both positive. The above system states precisely that, and actually any point in the first quadrant is a solution to the above system of inequalities.
8
(0, -14)
The solution to 40 plus 40 times 0 plus 1 is 41. The number name is forty-one.
That is not a correct equation. The numbers on the left equal -16, not 0. This is an equation with no solution.
x = 0
It is x = -5
9.46 + 0*(2+3+5+8) is the simplest solution.
Yes, 0, 2 is a solution to y=-x+2 because 0=-2+2 which can be rewritten as 2-2=0.
( 1 0 1 ) + ( 1 0 1 ) = 1 0 1 0