Systems of Equations

It's a set of equations that have more than one incognito, in which is necessary to find the values that meet the requirements from all equations.

1st Method: Replacement

Consists of choosing one of the equations and isolate the incognito to determinate it's value in relation to the other equation.

With that, the 2nd equation will be left with a single incognito and so we can find it's final value.

Example: Solve the following System of Equations

$$\begin{cases}x+y=12\\3x-y=20\end{cases}$$

Isolating the x on the first equation:

$$\begin{cases}x=12-y\\3x-y=20\end{cases}$$

so with x being equal to $$12-y$$ we can replace it on the second equation:

$$3(12-y)-y=20$$ $$36-3y-y=20$$ $$-4y=20-36$$ $$4y=16$$ $$y=\frac{16}{4}$$ $$y=4$$

finding y, we can replace the value on the first equation:

$$x+4=12$$ $$x=12-4$$ $$x=8$$

In this case, the result turns both equations true, since $$8+4=12$$ and $$3*8-4=20$$

Finally, the result can be represented as:

$$(8,4)$$

2nd Method: Addition

Here we join both equations into one, eliminating one of the incognitos.

For this it is necessary that the coefficients of the incognitos have the same value and opposing signs.

With opposing signs:

$$\begin{cases}x+y=12\\3x-y=20\end{cases}$$

In this case we already have this requirement met (+y on the first equation and -y on the second), with that we eliminate both y and sum the remaining values:

$$\frac{\begin{cases}x=12\\3x=20\end{cases}}{4x=32}$$ $$x=\frac{32}{4}$$ $$x=8$$

finding x we can now solve the equation and find y:

$$8+y=12$$ $$y=12-8$$ $$y=4$$

The result is the same of the first method:

$$(8,4)$$

Without opposing signs:

When the coefficients are not opposed, is necessary to multiply them by a number that transforms the coefficient in a opposing number to the other equation:

Example:

$$\begin{cases}3x+y=24\\5x+2y=60\end{cases}$$

In this case, since we cannot eliminate one of the incognitos, we need to multiply one equation by a number that transforms the coefficient into a opposite number from the other equation's coefficient. For example, we can try multiplying it to -2:

$$\begin{cases}3x+y=24\:(*-2)\\5x+2y=60\end{cases}=\begin{cases}-6x-2y=-48\\5x+2y=60\end{cases}$$ $$\frac{\begin{cases}-6x-2y=-48\\5x+2y=60\end{cases}}{-1x=12}$$

now we can do the rest like the other example:

$$x=-12$$ $$-6(-12)-2y=-48$$ $$72-2y=-48$$ $$-2y=-48-72$$ $$-2y=-120$$ $$y=\frac{120}{2}$$ $$y=60$$

The result can be written as:

$$(-12,60)$$

More than two equations

This topic can become more confusing when we add a third equation into the mix:

Example:

$$\begin{cases}3a+2b+5c=15\\7a-3b+2c=52\\5a+b-4c=2\end{cases}$$

In this case we can do two separate Systems of Equations: one with the first and second equation, and another one with the second and third. With that the result of both systems will be used to make a new one.

separating the system into two:

$$\begin{cases}3a+2b+5c=15\\7a-3b+2c=52\end{cases}\:\begin{cases}7a-3b+2c=52\\5a+b-4c=2\end{cases}$$

First system:

$$\begin{cases}3a+2b+5c=15\:(*2)\\7a-3b+2c=52\:(*5)\end{cases}=\begin{cases}6a+4b+10c=30\\-35a+15b-10c=-260\end{cases}$$

With a third incognito, we focus in eliminate c first:

$$\frac{\begin{cases}6a+4b=30\\-35a+15b=-260\end{cases}}{-29a+19b=-230}$$

Second system:

we do the same as in the first one:

$$\begin{cases}7a-3b+2c=52\:(*2)\\5a+b-4c=2\:(*1)\end{cases}=\begin{cases}14a-6b+4c=104\\5a+b-4c=2\end{cases}$$ $$\frac{\begin{cases}14a-6b+4c=104\\5a+b-4c=2\end{cases}}{19a-5b=106}$$

With that, we have two new equations, these can be solved in a new system to find a, b and consequentially c :

Third system

$$\begin{cases}-29a+19b=-230\\19a-5b=106\end{cases}$$ $$\begin{cases}-29a+19b=-230\:(*5)\\19a-5b=106\:(*19)\end{cases}=\begin{cases}-145a+95b=1150\\361a-95b=2014\end{cases}$$ $$\frac{\begin{cases}-145a+95b=1150\\361a-95b=2014\end{cases}}{216a=864}$$ $$a=\frac{864}{216}$$ $$a=4$$

replacing a in any equation in this third system, we have:

$$19(4)-5b=106$$ $$76-5b=106$$ $$-5b=106-76$$ $$b=\frac{30}{-5}$$ $$b=-6$$

in this case we use one of the first equations to find c :

$$3(4)+2(-6)+5c=15$$ $$12-12+5c=15$$ $$c=\frac{15}{5}$$ $$c=3$$

The result can be read as:

$$(4,-6.3)$$

Graphical Representation

We can also represent a equation system in a Cartesian Plane.

$$\begin{cases}x+y=4\\x-y=2\end{cases}$$

In each equation, we can attribute two values to one of the incognitos and then calculate the correspondent values to the other incognito, so that we obtain ordered pairs:

$$x+y=4$$

x	y	(x,y)
1	3	(1,3)
2	2	(2,2)

$$x-y=2$$

x	y	(x,y)
1	-1	(1,-1)
2	0	(2,0)

With that, we can insert these points in the Cartesian Plane:

By inserting the ordered pair resulted from the equation system on a graphical dynamic geometrical tool (in this case Geogebra), we can see that the point where both equations cross is actually the ordered pair that is the solution to this system.