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.

Example 5:

Solve the following system for $x$ and $y$ in terms of $a$ and $b$, where $%% a\neq b$.

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

Solution:

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


Substitution,


Elimination,


Matrices,


Graphing.

The Method of Substitution:

The method of substitution involves several steps:

Step 1:

Solve for $x$ in equation (1)

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

Step 2:

Substitute this value for $x$ in equation (2). This will change equation (2) to an equation with just one variable, $y$.

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

Step 3:

Solve for $y$ in the translated equation (2).

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

Step 4:

Substitute this value of $x$ in equation (1).

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

Step 5:

Check your answers by substituting the values of $x$ and $y$ 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.

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

The Method of Elimination:

The process of substitution involves several steps:

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.

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

Step 1:

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 $-a$ to form a new and equivalent equation 2.

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

Step 2:

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

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

Step 3:

Solve for y.

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

Step 4: Substitute this value of $y$ in the first equation and solve for x.

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

Step 5: Check your answers (See above check).

The Method of Matrices:

This method is essentially a shortcut for the method of elimination.

Rewrite equations (1) and (2) without the variables and operators. The left column contains the coefficients of the $x$'s, the middle column contains the coefficients of the $y$'s, and the right column contains the constants.

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

The objective is to reorganize the original matrix into one that looks like

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

where p and q are the solutions to the system.

Step 1.

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.

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

Step 2:

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.

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

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

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

Step 3:

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

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

Step 4:

Manipulate the matrix so that cell 12 is 0. Do this by adding

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

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

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

You can read the answers off the matrix as $x=\dfrac{1}{a^{2}+ab+b^{2}}$ and $y=\dfrac{a+b}{a^{2}+ab+b^{2}}$.

The method of Graphing:

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

If you would like to work a similar example, click on Example.

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

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Author: Nancy Marcus

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