Optional Donation Option 1 $1.00 USD Option 2$2.00 USD Option 3 $5.00 USD Option 4$10.00 USD Option 5 \$25.00 USD

## What is a system of equations?

A system of equations is a set of equations with a common solution.

Example 1:

$$\begin{eqnarray*} x^2 - 3 x + 2 &=& 0\\ x^2 - x - 2 &=& 0 \end{eqnarray*}$$

The first equation has solution $$x = 2$$ or $$x = 1$$, while the second has solution $$x = 2$$ or $$x = -1$$. The common solution is $$x = 2$$. Thus, that is the solution of the system.

## Solving a system of equations by substitution

Example 2:

$$\mathbb{S}_1$$:

$$\begin{eqnarray*} x + xy &=& 1\\ y - xy &=& - 4 \end{eqnarray*}$$

This system of equations has two unknown quantities, $$x$$ and $$y$$. The solution will consist of ordered pairs of numbers $$( x, y )$$ satisfying both equations.

A useful method for solving systems of equations is the method of substitution. In this method, one solves one of the equations in the system for a selected variable in terms of the other variables. Then all occurrences of the selected variable in the other equations of the system are replaced by the solution obtained.

In the example at hand, for the first equation in $$\mathbb{S}_1$$ (denoted $$E_1$$ ) the left side may be factored to yield $$x(1+y)=1$$. So neither $$x$$ nor $$1+y$$ may equal $$0$$. Thus we may divide both sides by $$1+y$$ to get the equivalent equation $$x = \dfrac{1}{ 1 + y }$$.

The $$x$$ in $$E_2$$ is replaced with $$\dfrac{1}{ 1 + y }$$ to produce an equation in only one variable, $$y$$. This yields a second system of equations which is equivalent to the first system of equations.

Substitution yields a second system $$\mathbb{S}_2$$ of equations $$E^\prime_1$$ and $$E^\prime_2$$ equivalent to the first.

$$\mathbb{S}_2$$:

$$\begin{eqnarray*} x &=& \dfrac{1}{ 1 + y }\\ y - \dfrac{y}{ 1 + y } &=& - 4 \end{eqnarray*}$$

Now, $$E^\prime_2$$ can be replaced by the equivalent equation $$( y + 2 )^2 = 0$$, which, in turn is equivalent to the equation $$y = - 2$$. Thus, the system

$$\mathbb{S}_3$$:

$$\begin{eqnarray*} x &=& \dfrac{1}{ 1 + y }\\ y &=& -2 \end{eqnarray*}$$

is equivalent to the original system of equations. Substituting the value $$y=-2$$ into the first equation yields $$x = -1$$ giving us the system of equations

$$\mathbb{S}_4$$:

$$\begin{eqnarray*} x &=& -1\\ y &=& -2 \end{eqnarray*}$$

which is equivalent to the original system. That means it has the same solution as the original system. But the solution of this last system is transparent. Thus, we have solved the original system.

In this example we see a general approach to solving all systems of equations. It consists of replacing one system by another equivalent system in a systematic way in order to produce (eventually) an equivalent system whose solution is obvious.

We have seen a couple of principles involved in replacing one system with an equivalent system:

1. Any equation in the system may be replaced by an equivalent equation. (The Equivalent Equation Principle)
2. Any variable or expression in an equation of the system may be replaced by another expression found to be equivalent to that variable or expression in one of the other equations of the system. (The Substitution Principle)

There are other principles which allow us to find equivalent systems, some of which will be discussed in the >.

## Exercise 7.1.1

Use the equivalent equation principle and the substitution principle to solve the system of equations:

$$\mathbb{S}_1$$:

$$\begin{eqnarray*} x + y &=& 0\\ x + y^2 &=& 0 \end{eqnarray*}$$

See Solution

## Exercise 7.1.2

Solve the system of equations:

$$\mathbb{S}_1$$:

$$\begin{eqnarray*} x + y &=& 1\\ 2x + y &=& -1 \end{eqnarray*}$$

See Solution