Solving Problems Using Elimination

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).

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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.

To create this article, volunteer authors worked to edit and improve it over time. This article can explain how to perform to achieve the solution for both variables.

Have you ever had a simultaneous problem equation you needed to solve?

Now, how do we know that a linear equation obtained by the addition of the first equation with a scalar multiplication of the second is equivalent to the first?

In this section we are going to solve systems using the Gaussian Elimination method, which consists in simply doing elemental operations in row or column of the augmented matrix to obtain its echelon form or its reduced echelon form (Gauss-Jordan).

This technique involves visually finding the point on the graph where the two lines cross to get the solutions for X and Y.

Because it helps you to graph quickly, having both equations in Y= form makes this method useful.

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.


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