Solving Problems Using Elimination

The bottom equation is then subtracted from the top one to cancel out a variable.This makes elimination efficient when the constants of both equations are already isolated.

The matrix we have obtained represents the system which is the solution to the initial system.

The augmented matrix of the system is We perform elemental operations in the rows to obtain the reduced row echelon form: We multiply the second row by 1/2 We add the first row with the second We multiply the first row by 1/3 This last equivalent matrix is in the reduced row echelon form and it has a null row, which means that the rows in the initial system are linearly dependent (either one of them can be obtained by another multiplying it by a scalar that is not null).

This method involves plugging an expression from one equation in for the variable in another.

To use this method, at least one variable in one of the equations must be isolated.

This can make the work take longer, and elimination is not the best choice in this scenario.

If the equations do not involve fractions or decimals, and you have a good visual understanding of linear equations, graphing on the coordinate plane is a good option.

Additionally, if the coefficients of the Xs or Ys in both equations are the same, elimination will get a solution quickly with minimal steps.

On the other hand, sometimes one or both whole equations have to be multiplied by a number to make the variable cancel.

The augmented matrix of the system is of the same dimension as the system (2x3).

The vertical line that separates the matrix coefficients from the vector of the independent terms.


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