Solution of a linear equation with two unknowns

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

Solution of a linear equation with two unknowns

In the discussion about linear equations we have seen that the equation $ax+b =0$ has solution $x=-\frac{b}{a}$ , at least, if $a\ne0$ . Herein $a$ and $b$ represent real numbers, for which we often have substituted concrete values. But as we have already done before, we can even write this equation and the solution with unknowns $a$ and $b$ in this general form. When we do that, then $a$ and $b$ are called parameters. In general, parameters are variables that occur in mathematical expressions, but do not play a role as unknowns in an equation. They represent distinghuised numbers that have not yet been specified.

It is however important that we mark $x$ as unknown. We could also have considered $a$ as unknown, in which case the solution would have been $a=-\frac{b}{x}$ , at least, when $x\ne0$ .

In this way we can also solve an equation with two unknowns.

The linear equation $ax+by=c$ , with unknowns $x$ and $y$ , in which $a$ , $b$ and $c$ are numbers or parameters, can be solved in the following two ways:

By considering $y$ temporarily as parameter: when we solve the linear equation with unknown $x$ , we get $x=-\frac{b}{a}y+\frac{c}{a}$ . This is only possible if $a\ne0$ ; the solutions are all dyads $[x,y]$ of the form $[-\frac{b}{a}y+\frac{c}{a},y]$ . Here, the role of $y$ as a parameter is clear: for each value of $y$ , there is exactly one solution.
considering $x$ temporarily as parameter: when we solve the linear equation with unknown $y$ , we get $y=-\frac{a}{b}x+\frac{c}{b}$ . This is only possible if $b\ne0$ ; the solutions are all dyads $[x,y]$ of the form $[x,-\frac{a}{b}x+\frac{c}{b}]$ . Here, the role of $y$ as a parameter is clear: for each value of $x$ , there is exactly one solution.

If $a\ne0$ and $b\ne0$ , then there is an overlap between the first and second case. Each of the two gives a way to describe the set of solutions. The first does so by considering $x$ as a function of $y$ , the second by considering $y$ as a function of $x$ . The vertical line at $a=0$ occurs exclusively in the first case, and the horizontal line at $b=0$ occurs only in the second case.

We now have identified all possible solutions when $a\ne0$ or $b\ne0$ . The results form the set of points on a line in the plane.
If $a=0$ and $b=0$ , then the equation is $0=c$ . So two cases are left:

if $c=0$ , then each dyad $[x,y]$ is a solution, and
if $c\ne0$ , then there is no solution.

Reduce $\;4 x+2 y=10\;$ to an equation of the form $\;y=a\,x+b$ .

$y = -2x + 5$

We proceed as in solving a linear equation with unknown $y$ . Thus, we consider $x$ as a parameter.

$\begin{array}{rclcl} 4 x+2 y&=&10&\phantom{xxxxx}&\blue{\text{the given equation}}\\ 2\cdot y &=& 10- 4 x&\phantom{xxxxx}&\blue{\text{movement of terms without }y\text{ to the right}}\\ y &=& \dfrac{10-4x}{2} &\phantom{xxxxx}&\blue{\text{division by the coefficient }2\text{ of }y} \\ y &=& -2 x + 5&\phantom{xxxxx}&\blue{\text{simplified order of terms}} \end {array}$

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