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