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.

Let's start with an example stated in narrative form. We'll convert it to an equivalent equation in algebraic form, and then we will solve it.

Example 3:

The sum of the digits in a two-digit number is 14. Furthermore, the number itself is 2 greater than 11 times the tens digits. Find the number.

Suppose the number is ab; it can be written as 10a + b, where a is the tens digit and b is the ones digit.

Solution:

We have two unknowns: the ones digit (b) and the tens digit (a) Our objective is to find these two numbers.

Sentence (1): The sum of the digits in a two-digit number is 14 can be restated as

$$\begin{eqnarray*} && \\ a+b &=&14 \\ && \end{eqnarray*}$$

Sentence (2): The number itself is 2 greater than 11 times the tens digits can be restated

$$\begin{eqnarray*} && \\ 10a+b &=&11a+2 \\ && \end{eqnarray*}$$

$$\begin{eqnarray*} (1) &:&a+b=14 \\ && \\ (2) &:&-a+b=2 \\ && \end{eqnarray*}$$

We have converted a narrative statement of the problem to an equivalent algebraic statement of the problem. Let's solve this system of equations.

     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 $b$ in equation (1).

$$\begin{eqnarray*} a+b &=&14 \\ && \\ b &=&14-a \\ && \end{eqnarray*}$$

Step 2:

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

$$\begin{eqnarray*} -a+b &=&2 \\ && \\ -a+\left( 14-a\right) &=&2 \\ && \\ && \end{eqnarray*}$$

Step 3:

Solve for a in the translated equation (2).

$$\begin{eqnarray*} && \\ -a+\left( 14-a\right) &=&2 \\ && \\ -a+14-a &=&2 \\ && \\ -2a &=&-12 \\ && \\ a &=&6 \\ && \end{eqnarray*}$$

Step 4:

Substitute this value of a in the b equation you obtained in Step 1.

$$\begin{eqnarray*} && \\ (1) &:&a+b=14 \\ && \\ 6+b &=&14 \\ && \\ b &=&8 \\ && \\ && \end{eqnarray*}$$

The number is 68.

Step 5: Check your answers by substituting the values of a and b 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.

$$\begin{eqnarray*} (1) &:&6+8=14 \\ && \\ (2) &:&10\left( 6\right) +8=11\left( 6\right) +2 \\ && \\ && \end{eqnarray*}$$

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.

Step 1:

Add the two equations.

$$\begin{eqnarray*} (1) &:&a+b=14 \\ && \\ (2) &:&-a+b=2 \\ && \\ 2b &=&16 \\ && \\ b &=&8 \\ && \\ && \end{eqnarray*}$$

Step 2:

Substitute $b=8$ in equation (1) and solve for $a$.

$$\begin{eqnarray*} && \\ a+8 &=&14 \\ && \\ a &=&6 \\ && \\ && \end{eqnarray*}$$

Step 3:

Check your answers in equation (2). Does

$$\begin{eqnarray*} && \\ 10\left( 6\right) +8 &=&11\left( 6\right) +2 \\ && \\ && \end{eqnarray*}$$

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 a's, the middle column contains the coefficients of the b's, and the right column contains the constants.

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ... 14 \\ & & & & \vert & & \\ -1 & & 1 & & \vert & & 2 \end{array} \right]$

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

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

where p and t 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, we don't have to do anything. The number 1 is already in the cell.

Step 2:

Manipulate the matrices so that the number in cell 21 (row2 - col1) is 0. Add Row 1 to Row 2 to form a new Row 2.

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

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ... 14 \\ & & & & \vert & & \\ 0 & & 2 & & \vert & & 16 \end{array} \right]$

Step 3:

Manipulate the matrix so that the cell 22 is 1. Do this by multiplying row 2 by 1/2.

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ...& 14 \\ & & & & \vert & & \\ 0 & & 1 & & \vert & & 8 \end{array} \right]$

Step 4: Manipulate the matrix so that cell 12 is 0. Do this by adding

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

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ... & 6 \\ & & & & \vert & & \\ 0 & & 1 & & \vert & & 8 \end{array} \right]$

You can read the answers off the matrix as $a=6,$ and $b=8,$ and the number is 68..

The method of Graphing:

In this method solve for b in each equation and graph both. The point of intersection is the 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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