SYSTEMS OF EQUATIONS in TWO VARIABLES

A system of equations is a collection of two or more equations with the same set of unknowns. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system.

The equations in the system can be linear or non-linear. This tutorial reviews systems of linear equations.

A problem can be expressed in narrative form or the problem can be expressed in algebraic form.

$\begin{eqnarray*} (1) &:&ax+by=\frac{1}{a} \\ && \\ (2) &:&b^{2}x+a^{2}y=1 \\ && \\ && \end{eqnarray*}$

A system of linear equations can be solved four different ways:

Substitution,

Elimination,

Matrices,

Graphing.

$\begin{eqnarray*} \left( 1\right) &:&ax+by=\frac{1}{a} \\ && \\ ax &=&\frac... ...1-aby}{a} \\ && \\ x &=&\frac{1-aby}{a^{2}} \\ && \\ && \end{eqnarray*}$

Substitute this value for

in equation (2). This will change equation (2) to an equation with just one variable,

$\begin{eqnarray*} && \\ \left( 2\right) &:&b^{2}x+a^{2}y=1 \\ && \\ \left... ...ht) &:&b^{2}\left( \frac{1-aby}{a^{2}}\right) +a^{2}y=1 \\ && \end{eqnarray*}$

$\begin{eqnarray*} && \\ (2) &:&b^{2}\left( \frac{1-aby}{a^{2}}\right) +a^{2}y... ... y &=&\frac{a^{2}-b^{2}}{a^{4}-ab^{3}} \\ && \\ && \\ && \end{eqnarray*}$

$\begin{eqnarray*} && \\ && \\ (1) &:&ax+by=\frac{1}{a} \\ && \\ && \\ ... ... && \\ && \\ x &=&\frac{1}{a^{2}+ab+b^{2}} \\ && \\ && \end{eqnarray*}$

Check your answers by substituting the values of

and

in each of the original equations. If, after the substitution, the left side of the equation equals the right side of the equation, you know that your answers are correct.

$\begin{eqnarray*} (1) &:&ax+by=? \\ && \\ a\left( \frac{1}{a^{2}+ab+b^{2}}\... ...ac{a^{2}+ab+b^{2}}{a^{2}+ab+b^{2}} &=&1 \\ && \\ && \\ && \end{eqnarray*}$

In a two-variable problem rewrite the equations so that when the equations are added, one of the variables is eliminated, and then solve for the remaining variable.

$\begin{eqnarray*} \left( 1\right) &:&ax+by=\frac{1}{a} \\ && \\ \left( 2\right) &:&b^{2}x+a^{2}y=1 \\ && \end{eqnarray*}$

Change equation (1) by multiplying equation (1) by $b^{2}$ to obtain a new and equivalent equation (1), and change equation (2) by multiplying equation (2) by

to form a new and equivalent equation 2.

$\begin{eqnarray*} && \\ \left( 1\right) &:&ax+by=\frac{1}{a} \\ && \\ &\r... ...+a^{2}y=1 \\ && \\ &\rightarrow &-ab^{2}x-a^{3}y=-a \\ && \end{eqnarray*}$

Add new equation (2) to new equation (1) to obtain equation (3). This should result in an equation with just one variable.

$\begin{eqnarray*} && \\ (1) &:&ab^{2}x+b^{3}y=\frac{b^{2}}{a} \\ && \\ (2... ...\frac{b^{2}}{a}-a=\frac{b^{2}-a^{2}}{a} \\ && \\ && \\ && \end{eqnarray*}$

$\begin{eqnarray*} && \\ \left( b^{3}-a^{3}\right) y &=&\frac{b^{2}-a^{2}}{a} ... ...y &=&\frac{a+b}{a\left( a^{2}+ab+b^{2}\right) } \\ && \\ && \end{eqnarray*}$

$\begin{eqnarray*} && \\ ax+by &=&\frac{1}{a} \\ && \\ ax+b\left[ \frac{a+... ... && \\ x &=&\frac{1}{a^{2}+ab+b^{2}} \\ && \\ && \\ && \end{eqnarray*}$

Rewrite equations (1) and (2) without the variables and operators. The left column contains the coefficients of the

's, the middle column contains the coefficients of the

's, and the right column contains the constants.

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ... & & & & \vert & & \\ b^{2} & & a^{2} & & \vert & & 1 \end{array} \right]$

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ... & & & & \vert & & \\ 0 & & 1 & & \vert & & q \end{array} \right] \bigskip$

Manipulate the matrix so that the number in cell 11 (row 1-col 1) is 1. In this case, Multiply row 1 by 1/a.

Manipulate the matrix so that the number in cell 21 is 0. To do this we rewrite the matrix by keeping row 1 and creating a new row 2 by adding $%% -b^{2}$ times row 1 to row 2.

$\begin{eqnarray*} && \\ -b^{2}\left[ Row\ 1\right] +\left[ row\ 2\right] &=&\left[ New\ Row\ 2\right] \\ && \\ && \end{eqnarray*}$

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ...ert & & -\dfrac{b^{2}}{a^{2}}+1 \end{array} \right] \bigskip\bigskip\bigskip$

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ...a^{3}-b^{3}}{a} & & \vert & & \dfrac{a^{2}-b^{2}}{a^{2}} \end{array} \right]$

Manipulate the matrix so that the cell 22 is 1. Do this by multiplying row 2 by $\dfrac{a}{a^{3}-b^{3}}$ .

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ...& \\ 0 & & 1 & & \vert & & \dfrac{a+b}{a^{2}+ab+b^{2}} \end{array} \right]$

$\begin{eqnarray*} && \\ -\frac{b}{a}[Row\ 2]+[Row\ 1] &=&[New\ Row\ 1] \\ && \end{eqnarray*}$

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ...& \\ 0 & & 1 & & \vert & & \dfrac{a+b}{a^{2}+ab+b^{2}} \end{array} \right]$

Since we don't know the values of a and b, we cannot use this method of solution.

If you would like to test yourself by working some problem similar to this example, click on Problem.