# How to solve the 3 equations with 2 variables?

## Question:

How to solve the 3 equations with 2 variables?

## Gaussian Elimination

Solving a system of linear equation, we apply Gaussian elimination (or use row reduction technique).

Gaussian elimination is a technique to solve a system of linear equations by eliminating the unknowns from the equations in such a manner that the matrix of the system is an upper triangular matrix.

A system is consistent if the augmented matrix has no pivot column in the last column.

A matrix with more rows than columns, will have a number of pivot columns equal or less than the number of columns.

The number of non-pivot columns gives the number of free parameters used in the infinite solutions.

A system with {eq}\displaystyle 2 {/eq} variables and {eq}\displaystyle 3 {/eq} equations, has an augmented matrix with at most {eq}\displaystyle 2 {/eq} pivot columns and at most {eq}\displaystyle 1 {/eq} non-pivot column.

So, the row echelon form of the system above is like

{eq}\displaystyle \begin{align} &\left[ \begin{array}{cc|c} a_{11}&a_{12}&b_1\\ 0&a_{22}&b_2\\ 0&0&b_{3}\\ \end{array} \right], \left[\text{ two pivot columns and consistent system if }b_3=0, \text{ therefore there is a unique solution } \right]\\ \text{ or }& \left[ \begin{array}{cc|c} a_{11}&a_{12}&b_1\\ 0&0&b_2\\ 0&0&b_{3}\\ \end{array} \right], \left[\text{ one pivot column and consistent system if }b_2=b_3=0, \text{ therefore there is are infinitely many solutions } \right]\\ \end{align} {/eq}

For example, the system of 3 equations with 2 unknowns, {eq}\displaystyle \begin{cases} x-y=1\\ 2x+y=2\\ y=0 \end{cases} {/eq}

has the augmented system given as {eq}\displaystyle \begin{align} \left[ \begin{array}{cc|c} 1&-1&1\\ 2&1&2\\ 0&1&0\\ \end{array} \right] \overset{-2R_1+R_2}{\iff} \left[ \begin{array}{cc|c} 1&-1&1\\ 0&3&0\\ 0&1&0\\ \end{array} \right]\overset{-\frac{1}{3}R_2+R_3}{\iff} \left[ \begin{array}{cc|c} 1&-1&1\\ 0&3&0\\ 0&0&0\\ \end{array} \right]. \end{align} {/eq}

So, there are two pivot columns and no no-pivot column, therefore there is a unique solution given by {eq}\displaystyle \begin{cases}x-y=1\\ 3y=0 \end{cases} \implies y=0 \text{ and } x=1, {/eq} there are infinitely many solutions.

Graphically, the solution is the point of intersection of three lines in plane.

But the system of 3 equations with 2 unknowns, {eq}\displaystyle \begin{cases} x-y=1\\ 2x+y=2\\ x+y=0 \end{cases} {/eq}

has the augmented system given as {eq}\displaystyle \begin{align} \left[ \begin{array}{cc|c} 1&-1&1\\ 2&1&2\\ 1&1&0\\ \end{array} \right] \overset{-2R_1+R_2}{\iff} \left[ \begin{array}{cc|c} 1&-1&1\\ 0&3&0\\ 1&1&0\\ \end{array} \right]\overset{-R_1+R_3}{\iff} \left[ \begin{array}{cc|c} 1&-1&1\\ 0&3&0\\ 0&0&1\\ \end{array} \right] \end{align} {/eq}

So, the system is inconsistent, because the last column of the augmented matrix is a pivot column, therefore there is no solution.

Graphically, the system is the set of three lines in plane that do not go all through the same point.

While the system {eq}\displaystyle \begin{cases} x-y=1\\ 2x-2y=2\\ -x+y=-1 \end{cases} {/eq} has the augmented system given as {eq}\displaystyle \begin{align} \left[ \begin{array}{cc|c} 1&-1&1\\ 2&-2&2\\ -1&1&-1\\ \end{array} \right] \overset{-2R_1+R_2}{\iff} \left[ \begin{array}{cc|c} 1&-1&1\\ 0&0&0\\ -1&1&-1\\ \end{array} \right]\overset{R_1+R_3}{\iff} \left[ \begin{array}{cc|c} 1&-1&1\\ 0&0&0\\ 0&0&0\\ \end{array} \right] \end{align} {/eq}

So, there is one pivot column and one non-pivot column, therefore there are infinitely many solution given by {eq}\displaystyle \begin{cases}x-y=1\\ y=t, t\in\mathbf{R} \end{cases} \implies \begin{cases}x=1+t\\ y=t \end{cases} {/eq}

Graphically, the system is the set of three identical lines in plane.