r + 2t = -3 . . . . (A)
3r - 4t = -9 . . . . (B)
2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 9
5r = - 15 so r = -3
substituting in (A), t = 0
So the answer is (r, t) = (-3, 0)
r + 2t = -3 . . . . (A)
3r - 4t = -9 . . . . (B)
2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 9
5r = - 15 so r = -3
substituting in (A), t = 0
So the answer is (r, t) = (-3, 0)
r + 2t = -3 . . . . (A)
3r - 4t = -9 . . . . (B)
2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 9
5r = - 15 so r = -3
substituting in (A), t = 0
So the answer is (r, t) = (-3, 0)
r + 2t = -3 . . . . (A)
3r - 4t = -9 . . . . (B)
2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 9
5r = - 15 so r = -3
substituting in (A), t = 0
So the answer is (r, t) = (-3, 0)
A system of equations is a set of two or more equations with the same variables, graphed in the same coordinate plane
The values for which the equations are solved. Graphically the intersection of the lines that are the solutions to the individual equations. The link below gives some explanations. The equations themselves will have to be given for a solution to be found.
No. There could be no solution - no values for x, y, and z so that the 3 equations are true.
Put the values that you find (as the solution) back into one (or more) of the original equations and evaluate them. If they remain true then the solution checks out. If one equation does not contain all the variables involved in the system, you may have to repeat with another of the original equations.
Depending on the context, the "feasible region" or "solution set".
The solution of a system of linear equations is a pair of values that make both of the equations true.
A system of equations is a set of two or more equations with the same variables, graphed in the same coordinate plane
The solution to a system on linear equations in nunknown variables are ordered n-tuples such that their values satisfy each of the equations in the system. There need not be a solution or there can be more than one solutions.
The values for which the equations are solved. Graphically the intersection of the lines that are the solutions to the individual equations. The link below gives some explanations. The equations themselves will have to be given for a solution to be found.
The values for which the equations are solved. Graphically the intersection of the lines that are the solutions to the individual equations. The link below gives some explanations. The equations themselves will have to be given for a solution to be found.
The values for which the equations are solved. Graphically the intersection of the lines that are the solutions to the individual equations. The link below gives some explanations. The equations themselves will have to be given for a solution to be found.
No. There could be no solution - no values for x, y, and z so that the 3 equations are true.
Put the values that you find (as the solution) back into one (or more) of the original equations and evaluate them. If they remain true then the solution checks out. If one equation does not contain all the variables involved in the system, you may have to repeat with another of the original equations.
This is the case when there is only one set of values for each of the variables that satisfies the system of linear equations. It requires the matrix of coefficients. A to be invertible. If the system of equations is y = Ax then the unique solution is x = A-1y.
Depending on the context, the "feasible region" or "solution set".
They make up the solution set.
In mathematics, a solution refers to a value or set of values that satisfies an equation, inequality, or system of equations. It is the value or values that make the equation or inequality true.