Solved examples of the System of Equations

System of equations: 

A system of equations is a collection of equations that are connected in such a way that they have a common solution. They are particularly useful in analysing situations where the solution depends on multiple factors.

For example consider the system of linear equations, 2x + 3y = 13 and 3x + 2y = 12

Here from the trial and error method we can say that the solution of this system of equations will have the solution, x = 2 and y = 3. But we need some standard procedures for finding such solutions, we will take a glimpse at some of those methods ahead.

Note: When we say a system of equations, it could mean a system of linear equations, non-linear equations, bilinear equations, polynomial equations, differential equations, etc.

Here we will study the system of linear equations. 

The solution that we get in analysing a system of linear equations will depend on the nature of the system of linear equations. It could be consistent with at least one solution or inconsistent with no solution.

For further study on a system of equations, we will check different methods to find the solution of the following system of equations.

3x + 4y + 3z = 37

5x + 2y +   z = 23

  x +   y + 4z = 26

Substitution method.

If we take a look at the equations above, we can see that there are 3 equations and 3 variables. In the substitution method, we reduce the number of variables from 3 to 1 by substituting for each variable and thereby bringing down the number of equations to 1 to find the solution. Now let’s solve this system of equations using the method of substitution.

3x + 4y + 3z = 37 ( equation 1 )

5x + 2y +   z = 23 ( equation 2 )

  x +   y + 4z = 26 ( equation 3 )

Here lets take a look at equation 2,in this equation we can represent z as,

z = 23 – ( 5x + 2y )

Now lets substitute this expression of z for z in equations 1 and 3,

For equation 1, 3x + 4y + 3( 23 – ( 5x + 2y ) ) = 37

⇒ 3x + 4y + 69 – 15x – 6y = 37

⇒ -12x – 2y = -32

⇒   6x +   y =  16 ( equation 4 )

For equation 3, x +   y + 4( 23 – ( 5x + 2y ) ) = 26

⇒ x +   y + 92 – 20x – 8y = 26

⇒ 19x + 7y = 66 ( equation 5 )

Now we have two equations with two variables, we now have a system of equations with two variables. We will use the same method to substitute for y now.

From equation 4,   y = 16 – 6x

We will now substitute the value of y in equation 5,

⇒ 19x + 7( 16 – 6x ) = 66

⇒ 19x – 42x = 66 – 112

⇒ -23x = -46

⇒ x = 2

Now substituting the value of x in equation 4 we get,

⇒   6×2 +   y =  16 

⇒   y = 16 – 6×2

⇒   y = 4

Now substituting the value of x and y in equation 2 we get,

⇒   5×2 + 2×4 +   z = 23

⇒   z = 5

So we now have the solution of the system of equations as x=2 , y = 4 , z = 5.

Elimination method.

In the elimination method, we eliminate one variable at a time from the system of equations by multiplying the equation by a constant and subtracting one equation from another. To understand this, let us solve a system of equations using the elimination method. Let us consider the same system of equation as above,

3x + 4y + 3z = 37 ( equation 1 )

5x + 2y +   z = 23 ( equation 2 )

  x +   y + 4z = 26 ( equation 3 )

We will now try to eliminate z from the system of equations, so we will now multiply equation 1 with 4, equation 2 with 12 and equation 3 with 3 so that the coefficient of z becomes equal in all the equations. This gives us the following system of equations,

12x + 16y + 12z = 148 ( equation 4 )

60x + 24y + 12z = 276 ( equation 5 )

  3x +   3y + 12z = 78 ( equation 6 )

We will now try to eliminate z from the equation by subtracting equation 5 from equation 4 and equation 6. Thus we get,

Equation 4 – equation 5 ⇒ -48x – 8y = -128

⇒   6x + y = 16 ( equation 7 )

Equation 6 – equation 5    ⇒ -57x – 21y = -198

  ⇒ 57x + 21y = 198 ( equation 8 )

Now we will use the same method of elimination to eliminate y from equation 7 and 8.

So we multiply equation 7 with 21 and equation equation 8 with 1 and subtract them, this gives us an equation of the form,

⇒ 21( 6x + y) – 1(57x + 21y) = 21×16 – 198

⇒ 69x = 138

⇒ x = 2

Now we can substitute the value of x in equation 7 to get,

6×2 + y = 16 ⇒ y = 4

Now we will substitute for both x and y in equation 3 to get,

2 + 4 + 4z = 26 ⇒ z = 5

So we now have the solution to the system of equations to be x=2 , y = 4 , z = 5, which is the same as in the above case.

Matrix method.

We will try to solve the system of equations using the matrix method. Here we will represent the coefficients of the variables in a 3×3 matrix. Let’s say M and the variables in a 3×1 matrix, let’s say X, and the constants in a 3×1 matrix, let’s say C. The relation between these matrices can be represented in the following way:

MX = C  ⇒

Now, if we multiply both sides of the equation with the inverse of M, we get

M-1MX = M-1C

⇒ X = M-1C

So, on multiplying M-1 with C, we get a 3×1 matrix which will give us the solution. So we now get,

This gives us the solution x=2 , y = 4 , z = 5.

Conclusion. 

We can use several methods to find the solution of a system of equations. For all those methods, we will get the same set of solutions. We can choose the method for finding the solution according to our convenience. We can use the method of elimination and substitution several times while solving the system of equations. We can use all three of these methods to find the solution to a system of equations for more than 3 variables.