# 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.

## 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:

$$ 2 \cdot 2 + 2 \cdot 3 = 10$$

$ 4 + 6 = 10$

$ 10 = 10$

## Systems of equations worksheets

**Graphing - simple** (6.3 MiB, 1,428 hits)

**Graphing - advanced** (6.4 MiB, 1,293 hits)

**Substitution - simple** (148.3 KiB, 1,978 hits)

**Substitution - advanced** (143.0 KiB, 951 hits)

**Systems of equations - Elimination - Simple** (344.3 KiB, 834 hits)

**Systems of equations - Elimination - Advanced** (380.8 KiB, 1,298 hits)