A linear equation with two unknowns

Solving linear equations and inequalities: An equation of a straight line in a plane

A linear equation with two unknowns

Suppose that $x$ represents the number $3$ and $y$ the number $2$ . Then, for example, holds $x+1=6-y$ . This means that $x=3, y=2$ satisfies the equation $x+1=6-y$ . For the numbers $2$ and $3$ you can write many more equations that they satisfy.

In practice, the situation is the other way around: $x$ and $y$ are unknown numbers that satisfy the equation $x+1=6-y$ and you are trying to find the possible values of $x$ and $y$ . In other words, you want to solve the equation. This is possible by reduction, i.e. by repeatedly writing an equivalent equation that is simpler than the previous one but has the same solution. In the given example, the equation can be reduced to $y=5-x$ and that means that for a arbitrary value for $x$ , say $x=a$ , the value of $y$ is given by $y=5-a$ . The isolation of the variable $y$ yields a linear formula, that is, a formula of the form $y=ax+b$ with numbers $a$ and $b$ .

The given example is of a special type: namely, it refers to a linear equation in $x$ and $y$ . In this section we focus on the case of a linear equation with two unknowns.

Let $x$ and $y$ be variables.

A linear equation with unknowns $x$ and $y$ is an equation that can be reduced through elementary operations to the basic form $ax+by+c= 0$ where $a, b,$ and $c$ are real numbers. We also speak of a linear equation of $x$ and $y$ , and a linear relationship between $x$ and $y$ .

There is no unique basic form: the equations $3x-2y+1=0$ and $6x-4y+2=0$ both have the basic shape, but they are different and can be carried over into one another through elementary operations.

With an elementary operation we mean expansion of brackets, regrouping of subexpressions, addition or subtraction of the same expression on either side of the equation, or multiplication or division by a nonzero number on both sides of the equation. We speak of a elementary reduction when all the steps in the reduction are elementary operations.

The expression to the left of the equal sign ( $=$ ) is called the left-hand side of the equation (above that $ax+ by + c$ ) and the expression on its right is called the right-hand side (above this is $0$ ).

The terms $ax, by$ and $b$ in the left-hand side of the basic form are called terms. The number $a$ is called the coefficient of $x$ and $b$ is the coefficient of $y$ . Terms that do not contain an unknown are called constant terms, or constants for short (above, these are the numbers $b$ and $0$ ).

A list of two numbers $[r,s]$ is called a solution of the equation if substitution of $x=r, y=s$ turns the equation into a true statement. All values $x,y$ in which the equation is true constitute the solution of the equation.

Two linear equations are called equivalent when they have the same solutions because they can be transformed into one another by elementary reduction.

If an equation can be reduced to another by elementary transformations, then the two equations are equivalent.

Substituting $b=0$ , $a=2$ and $c=3$ in the above basic form of a linear equation gives $2x+3=0$ . A solution of this equation is $x=-\frac{3}{2}$ . It is even the solution: there are no others.

We say that $x=-\frac{3}{2}$ is the solution of the equation $2x+3=0$ .

The equation $x+2=-(x+1)$ is an equivalent linear equation, and thus has the same solution.

The equations $x+y=xy$ and $x+\sqrt{y}=0$ are not linear because of the presence of the product $xy$ the square root respectively $\sqrt{y}$ .

The equations $x+y=x\cdot y$ and $x+\sqrt{y}=0$ are not linear, because the product $x\cdot y$ , respectively, the square root $\sqrt{y}$ is present.

Although the equation $\frac{x-1}{2x-3}=4\;$ and $|x-1|=|2x-3|\;$ can be solved using linear equations (check this yourself), we still cannot call them linear equations, as the left-hand side of the first equation is a rational expression and the second equation contains absolute values. As a result, there is no elementary reduction to the basic form of a linear equation.

The equation $(x-1)^2=(x+1)^2$ is considered linear, because expanding the brackets and regrouping terms lead to the equation $4x=0$ . It would be too artificial to consider this equation as a quadratic equation $0x^2+4x=0$ , just because there is a square in the original equation.

Conversely, we call the equation $\sqrt{(x-1)^2+4x}=0$ not linear, even though it can be reduced to $x+1=0$ . However, the reduction does contain a non-elementary step: the transition from $\sqrt{(x+1)^2}=0$ to $x+1=0$ .

Conversely, we call the equation $\sqrt{(x-1)^2+4x}=0$ not linear, even though it can be reduced to $x+1=0$ . However, the reduction does contain a non-elementary step: the transition from $\sqrt{(x+1)^2}=0$ to $x+1=0$ .

The concept of equivalence of equations may also be used for non-linear equations:

Two equations are equivalent if they have the same solution(s).

Each of the two non-linear equations $(x-1)^3=0$ and $(x-1)^2=0$ is equivalent to the linear equation $x-1=0$ . However, the first two equations are non-linear because, in order to reduce to the linear equation $x-1=0$ , a non-basic simplification is needed.

Non-elementary operation

Squaring both sides of an equation is an example of an operation that is not elementary and that does not always transform an equation into an equivalent one. For instance, squaring both sides of the equation $x=5$ gives $x^2=25$ , an equation having $x=-5$ as a solution next to $x=5$ . Besides, this operation transforms many linear equations into non-linear ones.

Reduce the equation $5x+3y-1=3x+6y+2$ to a basic form.

A possible basic form: $2x-3y-3=0$

This follows from the following reduction.
$\begin{array}{rclcl} 5x+3y-1&=& 3x+6y+2 &\phantom{x}&\blue{\text{the given equation}}\\5x+3y-1- 3x &=& 3x+6y+2 - 3x &\phantom{x}&\blue{\text{subtraction of }3x\text{ on both sides}}\\2x+3y-1 &=& 6y+2 &\phantom{x}&\blue{\text{simplification}}\\ 2x+3y-1- 6y &=& 6y+2 - 6y &\phantom{x}&\blue{\text{subtraction of }6y\text{ on both sides}}\\2x-3y-1 &=& 2 &\phantom{x}&\blue{\text{simplification}}\\2x -3y-1-2 &=& 2 -2&\phantom{x}&\blue{\text{subtraction of } 2\text{ on both sides}}\\2x-3y-3 &=& 0&\phantom{x}&\blue{\text{simplification}}\end{array}$ So a basic form of the linear equation is $2x-3y-3=0$ .

Another basic form you will get for instance dividing both sides of the newly acquired base form by $2$ , resulting in $x-{{3\cdot y}\over{2}}-{{3}\over{2}}=0$ .

New example