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Methods of solving systems of linear equations

substitution method

Methods of solving systems of linear equations

A systems of equations is a set of multiple equations that need to be solved. The goal is to find the unknown variable.

For example, we need a system of two equations to solve equations that have two variables. There arefour methods of solving systems of equations:

But first, lets remember:

It is valid:\forall a, b, c

  • if a<b, then \forall c\in \mathbb{R}:

a+c<b+c                                                   (1)

  • if a<b, then \forall c\in \mathbb{R}:

a \cdot c<b \cdot c                                 (2)

  •  if a<b and b<c

a<c                                                            (3)

Substitution method

Substitution is a simple method in which we solve one of the equations for one variable and then substitute that variable into the other equation and solve it.


\left\{\begin{array}{rcl} x+y=4  \\ x-y=2 \end{array} \right

To solve the system of two equations above we need to follow these steps:


\left\{\begin{array}{rcl} x+y=4  & \Rightarrow y=4-x \\ x-y=2 \end{array} \right


Now, we must solve this equation:



 we use property (1)

2x-4=2 &/+4


we use property (2)

2x=6 &/:2


Now, we must return value of unknown variable of in first equation: x+y=4

3+y=4 & /+(-3)




The result is: (x,y)=(3,1)

Now we can check our result.

\left\{\begin{array}{rcl} x+y=4  & \Rightarrow y=4-x \\ x-y=2 \end{array} \right


\left\{\begin{array}{rcl} 3+1=4   \\ 3-1=2 \end{array} \right

Let’s try to solve this system of equations using all methods.

Graphing method

If we want to solve system of linear equation by this method, we must know that each linear equation is the line. That’s the reason that we call that function linear function. Now, we can easily draw that lines in coordinate plane.

For the line x+y=5:

For the line x-y=1:

We can see that the result is (x,y)=(3,2). You can try to solve that problem by substitution method.

Addition method

\left\{\begin{array}{rcl} x+y=5   \\ x-y=1 \end{array} \right

+ \left\{\begin{array}{rcl} x+y=5   \\ x-y=1 \end{array} \right

2x=6 /:3


Now we return value of variable x in one of the equations and we get:

3+y=5 & /+(-3)



The result is of system od equations is: (x,y)=(3,2).


Gaussian elimination method

We can solve once again system of linear equation:

\left\{\begin{array}{rcl} x+y=5   \\ x-y=1 \end{array} \right

\left\{\begin{array}{rcl} x+y=5   \\ x-y=1 & /\cdot (-1) \end{array} \right

\left\{\begin{array}{rcl} x+y=5   \\ -x+y=-1 \end{array} \right

2y=4 /:2


Now,we can return variable y in equation.

x+2=5 /+(-2)



The result is of system of equations, once again, is: (x,y)=(3,2).


To check if the results are right lets put our results for variable x and variable y in the first equation:

\ 2x + 2y = 10
\ 2 \cdot 2 + 2 \cdot 3 = 10
\ 4 + 6 = 10
\ 10 = 10
We got the right results.

Systems of equations worksheets

  Graphing - simple (6.3 MiB, 827 hits)

  Graphing - advanced (6.4 MiB, 825 hits)

  Substitution - simple (148.3 KiB, 905 hits)

  Substitution - advanced (143.0 KiB, 521 hits)

  Systems of equations - Elimination - Simple (344.3 KiB, 236 hits)

  Systems of equations - Elimination - Advanced (380.8 KiB, 216 hits)