The system of equations generally refers to the system of linear equations. It is a set of linear equations (equations with degree one) which have same set of variables.

Step 1: Determine the value of any variable given in the system of equations in terms of other variables.
Step 2: Substitute this value into another equation.
Step 3: Calculate the value of that variable with the help of substitution.
Step 4: Find the value of other variables.
Let us take one example:
Solved Example
Question: Solve the following system of equations by substitution method.
$x+2y=1$
$2x+3y=5$
Solution:
$x+2y=1$
$2x+3y=5$
Solution:
Let us assign numbers to the given equations
$x+2y=1$ .......1
$2x+3y=5$ .....2
From equation 1, we get
$x=12y$ ........3
Substituting the value of x in equation 2,
$2(12y)+3y=5$
$24y+3y=5$
$2y=5$
$y=3$
Substitute the value of y in equation 3
$x=12(3)$
$x=7$
Hence, we have $x=7$ and $y=3$
$x+2y=1$ .......1
$2x+3y=5$ .....2
From equation 1, we get
$x=12y$ ........3
Substituting the value of x in equation 2,
$2(12y)+3y=5$
$24y+3y=5$
$2y=5$
$y=3$
Substitute the value of y in equation 3
$x=12(3)$
$x=7$
Hence, we have $x=7$ and $y=3$
Step 1: Make the coefficient of a particular variable in one equation equal to the coefficient of that variable in another equation. In order to do that, multiply with some number, so that the variable has same constant.
Step 2: Now, we have to eliminate this variable. In order to do that, we either have to add or subtract two equations and cancel out the variable whose constants are made equal. Thus, we get the value of another variable.
Step 3: Find the value of other variables by the same method.
Let us see an example.
Solved Example
Question: Determine the value of x and y from the following linear system
$3x+2y=6$
$x+4y=8$
Solution:
$3x+2y=6$
$x+4y=8$
Solution:
Let us allot numbers to the equations,
$3x+2y=6$ .......1
$x+4y=8$ .........2
We can see that the coefficient of x in the two equations are 3 and 1 respectively. In order to make them equal, we need to multiply the second equation by 3.
So, multiplying equation 2 by 3, we obtain,
$3x+12y=24$ .......3
Now, if we subtract equation 1 from equation 3, then variable x will be eliminated.
$3x+12y=24$
$3x2y=6$
3x and 3x cancel out and we are left with:
$10y=18$
$y$ = $\frac{9}{5}$
Further, multiplying equation 1 by 2, we get,
$6x+4y=12$ ........4
Now, subtract equation 2 from equation 4,
$6x+4y=12$
$x4y=8$
4y and 4y cancel out and we are left with:
$5x=4$
$x$ = $\frac{4}{5}$
Hence, we have the solution $y$ = $\frac{9}{5}$ and $x$ = $\frac{4}{5}$.
$3x+2y=6$ .......1
$x+4y=8$ .........2
We can see that the coefficient of x in the two equations are 3 and 1 respectively. In order to make them equal, we need to multiply the second equation by 3.
So, multiplying equation 2 by 3, we obtain,
$3x+12y=24$ .......3
Now, if we subtract equation 1 from equation 3, then variable x will be eliminated.
$3x+12y=24$
$3x2y=6$
3x and 3x cancel out and we are left with:
$10y=18$
$y$ = $\frac{9}{5}$
Further, multiplying equation 1 by 2, we get,
$6x+4y=12$ ........4
Now, subtract equation 2 from equation 4,
$6x+4y=12$
$x4y=8$
4y and 4y cancel out and we are left with:
$5x=4$
$x$ = $\frac{4}{5}$
Hence, we have the solution $y$ = $\frac{9}{5}$ and $x$ = $\frac{4}{5}$.
Step 1: Construct a table with two columns, one is for values of x and another is for values of y.
Step 2: Choose few values for x and substitute them in equation to find y. Write down all the values of x and y in the table.
Step 3: Graph the line on graph paper with the help of points so obtained.
Step 4: Similarly, follow the same process for the other equations too.
Step 5: The solution of the system of equations is the point where they intersect. Read coordinates of that point from graph. This is the required solution.
If the two lines do not intersect, it means that the lines are parallel to each other and hence they have no solution.
Solved Example
Question: Solve the following system of linear equations graphically:
$x+y=2$
$xy=4$
Solution:
$x+y=2$
$xy=4$
Solution:
$x+y=2$ .......1
$xy=4$ ........2
For equation 1,
We get the points (1, 3), (0, 2) (1, 1), (2, 0) and (3, 1)
For equation 2,
We get the points (1, 5), (0, 4) (1, 3), (2, 2) and (3, 1)
The graph of the above two equations is shown in the following figure:
We can easily see that the two lines intersect each other at point (3, 1). Therefore, the solution to the given system of equations is x = 3 and y = 1.
$xy=4$ ........2
For equation 1,
x  $y=2x$ 
1 
3 
0 
2 
1 
1 
2 
0 
3  1 
For equation 2,
x  $y=x4$ 
1 
5 
0 
4 
1 
3 
2 
2 
3  1 
The graph of the above two equations is shown in the following figure:
We can easily see that the two lines intersect each other at point (3, 1). Therefore, the solution to the given system of equations is x = 3 and y = 1.
The formula for a system of linear equations is:
X is the matrix of variables.
B is the matrix of constants.
The formula for solving system of equations is derived from the above formula and is given below:
Where, $A^{1}$ is the inverse of matrix A.
Formula for finding $A^{1}$ is given by:
Let us understand this with the help of an example:
Solved Example
Question: Solve following equations graphically
$x+y+z=6$
$2x+y+z=4$
$2x+2y+z=8$
Solution:
$x+y+z=6$
$2x+y+z=4$
$2x+2y+z=8$
Solution:
$x+y+z=6$ .....1
$2x+y+z=4$ .......2
$2x+2y+z=8$ .....3
$A=\begin{bmatrix}
1 &1 &1 \\
2 &1 &1 \\
2 &2 &1
\end{bmatrix}$
$B=\begin{bmatrix}
6\\
4\\
8
\end{bmatrix}$
$X=\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}$
Let us find $A^{1}$
$a_{11}=+(1 \times 11 \times 2)$ = 1
$a_{12}=(2 \times 11 \times 2)$ = 0
$a_{13}=+(2 \times 21 \times 2)$ = 2
$a_{21}=(1 \times 11 \times 2)$ = 1
$a_{22}=+(1 \times 11 \times 2)$ = 1
$a_{23}=(1 \times 21 \times 2)$ = 0
$a_{31}=+(1 \times 11 \times 1)$ = 0
$a_{32}=(1 \times 11 \times 2)$ = 1
$a_{33}=+(1 \times 11 \times 2)$ = 1
$M=\begin{bmatrix}
1 &0 &2 \\
1 &1 &0 \\
0 &1 &1
\end{bmatrix}$
Adjoint of matrix A is obtained by finding transpose matrix of M which is denoted by $M^{T}$ and is found by exchanging the elements of rows by that of columns.
Hence,
$adjA=M^{T}=\begin{bmatrix}
1 &1 &0 \\
0 &1 &1 \\
2 &0 &1
\end{bmatrix}$
Determinant of A = $\left  A \right =1(12)1(22)+1(42)$ = 1
$A^{1}$ = $\frac{adjA}{\left  A \right }$
$A^{1}=\begin{bmatrix}
1 &1 &0 \\
0 &1 &1 \\
2 &0 &1
\end{bmatrix}$
$X=A^{1}B$
$\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
1 &1 &0 \\
0 &1 &1 \\
2 &0 &1
\end{bmatrix}\begin{bmatrix}
6\\
4\\
8
\end{bmatrix}$
= $\begin{bmatrix}
6+4+0\\
04+8\\
12+08
\end{bmatrix}$
= $\begin{bmatrix}
2\\
4\\
4
\end{bmatrix}$
So, we have the solution set x = 2, y = 4 and z = 4.
$2x+y+z=4$ .......2
$2x+2y+z=8$ .....3
$A=\begin{bmatrix}
1 &1 &1 \\
2 &1 &1 \\
2 &2 &1
\end{bmatrix}$
$B=\begin{bmatrix}
6\\
4\\
8
\end{bmatrix}$
$X=\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}$
Let us find $A^{1}$
$a_{11}=+(1 \times 11 \times 2)$ = 1
$a_{12}=(2 \times 11 \times 2)$ = 0
$a_{13}=+(2 \times 21 \times 2)$ = 2
$a_{21}=(1 \times 11 \times 2)$ = 1
$a_{22}=+(1 \times 11 \times 2)$ = 1
$a_{23}=(1 \times 21 \times 2)$ = 0
$a_{31}=+(1 \times 11 \times 1)$ = 0
$a_{32}=(1 \times 11 \times 2)$ = 1
$a_{33}=+(1 \times 11 \times 2)$ = 1
$M=\begin{bmatrix}
1 &0 &2 \\
1 &1 &0 \\
0 &1 &1
\end{bmatrix}$
Adjoint of matrix A is obtained by finding transpose matrix of M which is denoted by $M^{T}$ and is found by exchanging the elements of rows by that of columns.
Hence,
$adjA=M^{T}=\begin{bmatrix}
1 &1 &0 \\
0 &1 &1 \\
2 &0 &1
\end{bmatrix}$
Determinant of A = $\left  A \right =1(12)1(22)+1(42)$ = 1
$A^{1}$ = $\frac{adjA}{\left  A \right }$
$A^{1}=\begin{bmatrix}
1 &1 &0 \\
0 &1 &1 \\
2 &0 &1
\end{bmatrix}$
$X=A^{1}B$
$\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
1 &1 &0 \\
0 &1 &1 \\
2 &0 &1
\end{bmatrix}\begin{bmatrix}
6\\
4\\
8
\end{bmatrix}$
= $\begin{bmatrix}
6+4+0\\
04+8\\
12+08
\end{bmatrix}$
= $\begin{bmatrix}
2\\
4\\
4
\end{bmatrix}$
So, we have the solution set x = 2, y = 4 and z = 4.
Solved Examples
Question 1: Sum of the digits of a twodigit number is 5. The number obtained by reversing the places of digits is 9 bigger than the original number. What was the original number?
Solution:
Solution:
Let the digit at ones place = x
and the digit at tens place = y
Then, the number will be represented as x + 10y
The number obtained after reversing the order of digits = 10x + y
According to the question:
Sum of the digits $x+y=5$ ................1
and
$10x+y=x+10y+9$
$10xx+y10y=9$
$9x9y=9$
$xy=1$ ................2
Now, solving the obtained equations 1 and 2 by substitution method,
From equation 1:
$x=5y$
Plugging this value in equation 2:
$5yy=1$
$y=2$
Plugging the value of y in equation 1:
$x+2=5$
$x=3$
Thus, the original number = x + 10y
= 3 + 10 * 2 = 23
and the digit at tens place = y
Then, the number will be represented as x + 10y
The number obtained after reversing the order of digits = 10x + y
According to the question:
Sum of the digits $x+y=5$ ................1
and
$10x+y=x+10y+9$
$10xx+y10y=9$
$9x9y=9$
$xy=1$ ................2
Now, solving the obtained equations 1 and 2 by substitution method,
From equation 1:
$x=5y$
Plugging this value in equation 2:
$5yy=1$
$y=2$
Plugging the value of y in equation 1:
$x+2=5$
$x=3$
Thus, the original number = x + 10y
= 3 + 10 * 2 = 23
Question 2: Determine the value of two digits, provided that sum of the digits is 10 and difference of the digits is 6.
Solution:
Solution:
Let us assume two digits to be x and y.
According to the conditions given in question:
$x+y=10$ ...........1
$xy=6$ ..............2
Solving above equations by elimination method:
Adding equation 1 and equation 3, we have
$x+y=10$
$xy=6$
y and y cancel out and we are left with:
$2x=16$
$x=8$
Further, subtracting equation 2 from equation 1,
$x+y=10$
$x+y=6$
x and x cancel out and we are left with:
$2y=4$
$y=2$
Hence, the two digits are 8 and 2 whose sum is 10 and difference is 6.
According to the conditions given in question:
$x+y=10$ ...........1
$xy=6$ ..............2
Solving above equations by elimination method:
Adding equation 1 and equation 3, we have
$x+y=10$
$xy=6$
y and y cancel out and we are left with:
$2x=16$
$x=8$
Further, subtracting equation 2 from equation 1,
$x+y=10$
$x+y=6$
x and x cancel out and we are left with:
$2y=4$
$y=2$
Hence, the two digits are 8 and 2 whose sum is 10 and difference is 6.