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 several ways.

Example 2:

How much of a 7% acid solution should be mixed with how much of a 15% acid solution so that you obtain 20 gallons of a 12% acid solution?

A 7% acid solution means that 0.07 times the total solution is acid.

Solution:

We have two unknowns: the amount of 7% acid solution and the amount of 15% acid solution. Our objective is to find these two numbers.

The sentence essentially says that if we add a certain amount of the 7% percent acid solution to a certain amount of the 15% acid solution, we will have 20 gallons with a 12% acid solution. Let's restate this sentence with two sentences.

Sentence (1):''The amount of the 7% acid solution'' + ''the amount of the 15% acid solution'' = 20 gallons.

Sentence (2): ''The amount of acid in the 7% acid solution'' + ''the amount of acid in the 15% acid solution'' = 0.12(20) = 2.4 gallons.

It is going to get tiresome writing the two phrases (the amount of the 7% acid solution) and (the amount of the 15% acid solution) over and over again. So let's write them in shortcut form. Call the phrase (the amount of 7% acid solution) by the symbol $x$ and call the phrase (the amount of 15% acid solution) by the symbol $y$.

The amount of acid in the 7% acid solution can then be written as $0.07x$, and the amount of acid in the 15% acid solution can be written as $0.15y$.

Let's rewrite sentences (1) and (2) in shortcut form.

$$\begin{eqnarray*} (1) &:&x+y=20 \\ && \\ (2) &:&0.07x+0.15y=0.12(20)=2.40 \\ && \end{eqnarray*}$$

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 y in equation (1).

$$\begin{eqnarray*} x+y &=&20 \\ && \\ y &=&20-x \\ && \end{eqnarray*}$$

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

$$\begin{eqnarray*} 0.07x+0.15y &=&2.40 \\ && \\ 0.07x+0.15\left( 20-x\right) &=&2.40 \\ && \\ && \end{eqnarray*}$$

Step 3: Solve for x in the translated equation (2).

$$\begin{eqnarray*} && \\ 0.07x+0.15\left( 20-x\right) &=&2.40 \\ && \\ 0.0... ...2.40 \\ && \\ -0.08x &=&-0.60 \\ && \\ x &=&7.5 \\ && \end{eqnarray*}$$

Step 4: Substitute this value of x in equation (1) to solve for y.

$$\begin{eqnarray*} && \\ (1) &:&x+y=20 \\ && \\ 7.5+y &=&20 \\ && \\ y &=&12.5 \\ && \\ && \end{eqnarray*}$$

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.

$$\begin{eqnarray*} (1) &:&x+y=20 \\ && \\ 7.50+12.50 &=&20 \\ && \\ && \... ...5\right) +0.15\left( 12.50\right) =2.40 \\ && \\ && \\ && \end{eqnarray*}$$

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.

$$\begin{eqnarray*} (1) &:&x+y=20 \\ && \\ (2) &:&0.07x+0.15y=2.40 \\ && \end{eqnarray*}$$

Step 1: Change equation (1) by multiplying both sides of equation (1) by $-0.07$ to obtain a new and equivalent equation (1).

$$\begin{eqnarray*} (1) &:&x+y=20 \\ && \\ (New\ 1) &:&-0.07x-0.07y=-1.4 \\ && \\ && \end{eqnarray*}$$

Step 2: Add new equation (1) to equation (2) to obtain equation (3) and solve for $y$.

$$\begin{eqnarray*} && \\ (New\ 1) &:&-0.07x-0.07y=-1.40 \\ && \\ (2) &:&0.... ... && \\ (3) &:&0.08y=1 \\ && \\ y &=&12.50 \\ && \\ && \end{eqnarray*}$$

Step 3: Substitute $y=12.50$ in equation (1) and solve for x.

$$\begin{eqnarray*} && \\ (1) &:&x+y=20 \\ && \\ x+12.50 &=&20 \\ && \\ x &=&7.50 \\ && \\ && \end{eqnarray*}$$

Step 4: Check your answers in equation (2).

$$\begin{eqnarray*} && \\ (2) &:&0.07x+0.15y=2.40 \\ && \\ &&0.07\left( 7.50\right) +0.15\left( 12.50\right) =2.40 \\ && \\ && \\ && \end{eqnarray*}$$

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 x's, the middle column contains the coefficients of the y's, and the right column contains the constants.

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ... & & & & \vert & & \\ 0.07 & & 0.15 & & \vert & & 2.40 \end{array} \right]$

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

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

where a and b 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 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 $%% -0.07\ \times \ row\ 1\ to\ row\ 2$.

$$\begin{eqnarray*} && \\ -0.07\left[ Row\ 1\right] +\left[ row\ 2\right] &=&\left[ New\ Row\ 2\right] \\ && \end{eqnarray*}$$

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ...0 \\ & & & & \vert & & \\ 0 & & 0.08 & & \vert & & 1 \end{array} \right]$

Step 3: Manipulate the matrix so that the cell 22 is 1. Do this by multiplying row 2 by 12.50.

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ... \\ & & & & \vert & & \\ 0 & & 1 & & \vert & & 12.50 \end{array} \right]$

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

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

$\begin{array}{r} (1) \\ \\ (2) \end{array} \left[ \begin{array}{rrrrr... ... \\ & & & & \vert & & \\ 0 & & 1 & & \vert & & 12.50 \end{array} \right]$

You can read the answers off the matrix as $x=7.50$ gallons and $y=12.50$ gallons.

The method of Graphing:

In this method, solve for y 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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