### Solving linear equations and inequalities: Systems of linear equations in two unknowns

### Solving systems of equations by the substitution method

The most obvious method of solving two equations with two unknown is the following.

Substitution method

1. Solve \(x\) with \(y\) as a parameter from one of the two equations.

\(\phantom{1.}\)This gives an expression of \(x\) in terms of \(y\).

2. Substitute the term wherever \(x\) is in the other equation.

\(\phantom{2.}\)Then a linear equation with the only unknown \(y\) emerges.

3. Solve this equation. As a result, \(y\) is determined.

4. Finally, determine \(x\) by substituting the found value of \(y\) in the equation that begins with \(x=\).

Solve the following system of equations: \[\left\{\;\begin{aligned}2x+3y &=1 \\ 5x+7y&=3\end{aligned} \right.\]

- We use the first equation to express \(x\) in \(y\): \[x = \frac{1}{2}-\frac{3}{2}y\]
- We substitute the expression for \(x\) in the second equation: \[5\left(\frac{1}{2}-\frac{3}{2}y\right)+7y = 3\]
- We solve this equation with unknown \(y\): Expanding brackets leads to \[\frac{5}{2}-\frac{1}{2}y=3\] So \(-\tfrac{1}{2}y = \tfrac{1}{2}\) , or \(y=-1\) .
- We find the value for \(x\) by substituting \(y=-1\) in the outcome of step 1: \[ x=-\tfrac{3}{2}\times 1+\tfrac{1}{2}= \tfrac{4}{2}=2\]

Thus, the solution of the system of equations is \(x=2\) and \(y=-1\), also denoted as \[x=2\quad\land\quad y=-1\]

There is not always exactly one solution; Two cases can be distinguished:

- The two equations can be dependent, that is to say that one equation is a multiple of the other. This case reveals itself in step 2: if we eliminate \(x\) from the equation, it is possible that \(y\) disappears. If the equation becomes \(0=0\), then it does not limit the solution set and we can ignore this equation. There remains one equation with two unknowns, which is the equation of a straight line.
- The equations may contradict each other, in the sense that no solution of one equation is also a solution of the other. This case also reveals itself in step 2: if we eliminate \(x\) from the equation, it is possible that \(y\) disappears (as in the previous case). But this time, the equation becomes \(0=c\) for a number \(c\) unequal to zero. This equation is never satisfied. So, the answer is that the system of equations has no solutions. We speak in this case of an
**inconsistent system of equations.**

In summary, there are three possibilities for a system of two linear equations with two unknowns:

- exactly one solution
- infinitely many solutions, which together represent points on a straight line
- no solution

For those who want to study more examples:

&&&\phantom{xxx}&\blue{\text{so that }x\text{ occurs in the first equation}}\\

x ={\displaystyle -{{6 y}\over{7}}-{{2}\over{7}} }& \land &{\displaystyle -x-y=4 } &\phantom{xxx}&\blue{x\text{ expressed in }y\text{ in the first equation}}\\ \\

x ={\displaystyle -{{6 y}\over{7}}-{{2}\over{7}} } & \land & {\displaystyle {{2}\over{7}}-{{y}\over{7}}=4 } &\phantom{xxx}&\blue{x\text{ replaced by the expression in }y}\\

&& &\phantom{xxx}&\blue{\text{from the first equation in the second }}\\

x ={\displaystyle -{{6 y}\over{7}}-{{2}\over{7}}} &\land & y = -26 &\phantom{xxx}&\blue{\text{The second equation solved}}\\ \\

x = 22 &\land & y = -26 &\phantom{xxx}&\blue{\text{The obtained value of }y\text{ substituted}}\\

& & &\phantom{xxx}&\blue{\text{into the first equation}}\\ \end{array}

\]

The answer is \[x= {\displaystyle 22}\land y = {\displaystyle -26}\]