Following is a system of linear equations of two variables, $x$ and $y$,
$2x-y=3$
$3x+2y=8$
Solution of a system of equations
In each of the above equations, let us substitute, $x=2$ and $y=1$.
First equation:
$2\cdot 2-1=3$
Simplifying,$3=3$
This is a true statement as the left hand side is equal to the right hand side. Therefore, $x=2$ and $y=1$, are the solution of the first equation.
Second equation:
$3\cdot 2+2\cdot 1=8$
Simplifying,$8=8$
This is also a true statement as the left hand side is equal to the right hand side. Therefore, $x=2$ and $y=1$ are also the solution of the second equation.
Since $x=2$ and $y=1$ are the solution of both the equations of the system, these are the solution of the system. We write the solution as ordered pair as $(x,y)$, first, the $x$ value, then the $y$ value. Therefore, the solution of the system as ordered pair is $(2,1)$.
Note that the solution of a system of equations, is the solution of all the equations in the system. If the solution of one equation is not the solution of the other equation, then that is not the solution of the system.
There are different methods by which you can find the solution of a system of equations, such as graphical method, substitution method, and eliminations method.
Graphical method
In the figure below, I have drawn the graphs of the system of equations,$\begin{align*} 2x-y&=3\\ 3x+2y&=8 \end{align*}$
Note that, not all the system of equations have a solution. Some have, some not and some have infinite many solutions.
If the graphs of the equations of a system are parallel, i.e., they do not intersect, then the system has no solution.
If the graphs overlap, then the system has infinite number of solutions. In this case, the equations are dependent, i.e, the two equations are the same but written in different forms.
Example 1: Solve the following system of equations,
$\begin{align*} -x-2y&=-4\\ x+2y&=2 \end{align*}$The graphs for these two equations are in the figure below.
Example 2:Solve the system of equations,
$\begin{align*} -2y=x-4\\ x+2y=4 \end{align*}$The graphs of the above equations are in the figure below.
Substitution method
Now, let us take the following system of equations,$\begin{align*} x+4y&=-1\\ 2x-3y&=9 \end{align*}$
and solve them to find the solution in another method, called substitution method as follows.
Take one of the equation, (I take the first equation) and solve for one variable, I take the variable $x$ as it is easy to solve for:
$x=-1-4y$
Now, substitute this $x$ in the second equation, you will get
$2(-1-4y)-3y=9$
Solve this equation for $y$,$-2-8y-3y=9$
Simplifying,$-2-11y=9$
Add $+2$,$-11y=11$
Divide $-11$
$y=-1$
Next, substitute this $y$ value in the equation that we solved for $x$: $x=-1-4y$,$x=-1-4\cdot (-1)$
Simplifying,$x=3$
Now you have the $x$ and the $y$ values, write the solution as ordered pair. The solution is $(3,-1)$.
If a system has no solution or infinite many solutions, then in the second step, the variable will get canceled and you will end up with a number on the left hand side and another number on the right hand side. If the numbers are not the same, i.e., the statement is not true, then the system has no solution. And if the numbers are the same, i.e., the statement is true, then the system has an infinite number of solutions.
Example 1: Solve the following system of equations
$\begin{align*} 5x+4 &=y\\ 5x-y &=-8\\ \end{align*}$Solution:
In the first equation, we already have $y$, so we do not have to solve for that. Just substitute $y$ in the second equation,
$5x-(5x+4)=-8$
Simplifying,$-4=-8$
This statement is not true, therefore, the system of equations has no solution.A system of equations with no solution is called inconsistent system.
The elimination method
There is another method, called elimination method that you can use to solve a system of linear equations.Let us take the following system of equations and solve them by this method.
$\begin{align*} x+5y&=7\\ 3x-2y&=4 \end{align*}$In the elimination method, you will eliminate one of the variable by adding the two equations. For that, first you need to make sure both the equations are in the standard form, $(ax+by=c)$. If one or both the equations are not in the standard form, write the equation(s) in standard form, first. Next, you need to make the coefficient of one of the variable in one equation is opposite to the coefficient of the same variable in the other equation, by multiplying with a number.
By looking at the system of equations, both of them are in standard form. Next, you need to make the coefficient of one of the variables in one equation is opposite of the coefficient of the same variable in the other equation. To do that first look at the equations, and see if the coefficients of any one of the variables are opposite. If not, look if the coefficients are the same (If the coefficients are the same, then you need to multiply, $-1$ with one of the equation). If not, look for the variable with coefficient 1. Here, in the first equation, the coefficient of the variable $x$ is $1$. Now, look for the coefficient of $x$ in the other equation. The coefficient of $x$ in the second equation is 3, so we need to have a coefficient of $-3$ for the $x$ in equation 1 to have an opposite. So, we multiply a $-3$ with the first equation,
$-3\times Eqn.1:$
$\:-3x-15y=-21 $
Now add this equation with the other equation, (Eqn. 2)
$\begin{aligned} -\cancel{3x}-15y &= -21 \\ \cancel{3x}-2y &= 4\\ \\ -17y &= -17\\ \end{aligned}$
Dividing, $-17$,
$y= 1$
Now, substitute this $y$ value to one of the original equation (any equation) and solve for $x$. I take the equation 1, and substitute $y=1$:
$x+5\cdot 1=7$
Solving for $x$,$x=2$
So, the solution is $(2,1)$.
Note that if an equation has no solution or infinite number of solutions, then when you eliminate one of the variable, both the variables will be eliminated. And, you will end up with a number on the left hand side and another number on the right hand side. If the two numbers are the same, then the system has no solution and if they are different, then the system has infinite number of solutions.
Example 1: Solve the following system of equations by elimination method:
$\begin{align*} 6y =12-4x \\ 2x+3y = 2\\ \end{align*}$Here, the equation 1 is not in the standard form, add $4x$ on both sides to make the equation in the standard form:
$4x+6y=12$
This equation has a common factor of $2$. Divide the equation by this common factor.
$2x+3y=4$
Now, we need to solve these equations:
$\begin{align*} 2x+3y = 4 \\ 2x+3y = 2\\ \end{align*}$The coefficients of all the variables are the same. So, we just multiply a $-1$ with one of the equation, I take the first equation, and multiply with $-1$, this will change the sign of all the terms of the equation:
$-2x-3y=-4$
Now, add this with the second equation:
$\begin{align*} -\cancel{2x}-\cancel{3y} &=-4\\ \cancel{2x}+\cancel{3y} &= 2\\ \\ 0 &=4 \end{align*} $
This statement is not, true, therefore, the system of equation has no solution.
Example 2:Solve the system
$\begin{align*} -2x+3y &= 6 \\ 8x-12y &= -24\\ \end{align*}$First make sure, the equations are in the standard form. Yes, they are.
The second equation has a common factor of $4$. Dividing out the common factor in the second equation,
$2x-3y=-6$
Now, we have these equations to solve:
$\begin{align*} -2x+3y &= 6 \\ 2x-3y &= -6\\ \end{align*}$Now, we see that the coefficients of $x$ in the equations are opposite and also the coefficients of $y$ are opposite. Add the two equations:
$\begin{align*} -\cancel{2x}+\cancel{3y} &= \cancel{6} \\ \cancel{2x}-\cancel{3y} &= -\cancel{6}\\ \\ 0=0 \end{align*}$This statement is true, so, the equation has an infinite number of solution.
A system of equations that has infinite number of solutions is called dependent. A dependent system has both the equations are the same but written in different form.