# Methods of solving systems of linear equations

A system of linear equations is a set of multiple linear equations. The goal is to find the ordered  $n$- tuple which satisfies all $n$ equations. Therefore, that ordered $n$ – tuple is the solution of that system. For example, ordered pair $(x,y)$ which satisfies both equations of the system is a solution of that system.

In order to solve equations that have two variables, we need a system of two equations. There are four methods of solving systems of equations.

But first, lets remember:

It is valid:  $\forall a, b, c \in \mathbf{R}$

• 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.

Example 1:

Solve the following system of linear equations using substitution method:

$$x+y=4$$

$$x-y=2$$

Solution:

To solve the system of two equations above we need to follow these steps. For example, let’s solve the first equation for $y$:

$$x+y=4 \Rightarrow y=4-x$$

$$x-y=2$$

_______________________

Now, we can substitute the expression $y=4-x$ in for $y$ in the second equation of our system:

$x-(4-x)=2$

$x-4+x=2$

We use property (1):

$$2x-4=2 /+4$$

$$2x-4+4=2+4$$

We use property (2):

$$2x=6 / \cdot \frac{1}{2}$$

$x=3$

Now, we must return the value $x=3$ in the first equation: $x+y=4$

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

$3+y-3=4-3$

$y=1$

The result is ordered pair $(x,y)=(3,1)$.

Now we can check our result. As we have already mentioned, the ordered pair which is the solution of the system, must sastisfy both equations.

$$x+y=4$$

$$x-y=2$$

________________

$$3+1=4 \Rightarrow 4=4$$

$$3-1=2 \Rightarrow 2=2$$

Now we are sure that $(x,y)=(3,1)$ is the solution of the given system.

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 the graph of 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 $x-y=1$:

The solution of the system is a point in which those two lines intersect. We can see that the result is $(x,y)=(3,2)$.

Now we can try to solve the same problem by addition method.

This method is sometimes also called opposite coefficients method. If the coefficients of $x$ (or $y$) are already opposite numbers, we just need to add the two equations to eliminate $x$ (or $y$). In contrary, we must achieve that those coefficients are opposites so that summation of the equations eliminates them.

Example 2:

Solve the following system of linear equations using addition method:

$$x+y=5$$

$$x-y=1$$

________

Solution:

We notice that the coefficients of $y$ (numbers $1$ and $-1$) are already opposites, so we just add the equations:

$$x+y=5$$

$$x-y=1$$

________

$$2x=6 /:2$$

$$x=3$$

Now we return value of variable $x$ in one of the equations, for example in the first equation, and we get:

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

$$3+y-3=5-3$$

$$y=2$$

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

Example 3:

Solve the following system of linear equations using addition method:

$$2x+3y=-1$$

$$3x-4y=-10$$

Solution:

In this example none of the coefficients are opposites. Let’s try to achieve opposite coefficients of variable $x$:

We can multiply the first equation by number $-3$:

$$-6x-9y=3$$

We can multiply the second equation by number $2$:

$$6x-8y=-20$$

Now we can add the equations:

$$-17y=-17$$

$$y=1$$

We must return the value $y=1$ in one of the equations:

$$2x+3 \cdot 1 = -1$$

$$2x=-4$$

$$x=-2$$

We can check our result:

$$2 \cdot (-2) + 3 \cdot 1 = -1 \Rightarrow -1=-1$$

$$3 \cdot (-2) – 4 \cdot 1 = -10 \Rightarrow -10=-10$$

Therefore, the solution is: $(-2,1)$.

## Gaussian elimination method

We can solve once again system of linear equation:

$$x+y=5$$

$$x-y=1$$

___________

$$x+y=5$$

$$x-y=1 /\cdot (-1)$$

________

$$x+y=5$$

$$-x+y=-1$$

$$2y=4 /:2$$

$$y=2$$

Now,we can return variable $y$ in equation.

$$x+2=5 /+(-2)$$

$x+2-2=5-2$

$x=3$

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, 1,300 hits)

Graphing - advanced (6.4 MiB, 1,199 hits)

Substitution - simple (148.3 KiB, 1,603 hits)

Substitution - advanced (143.0 KiB, 848 hits)

Systems of equations - Elimination - Simple (344.3 KiB, 684 hits)

Systems of equations - Elimination - Advanced (380.8 KiB, 760 hits)