Solving a System of Linear Equations

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Introduction
A system of linear equations (or linear system) is a group of (linear) equations that have more than one unknown factor. The unknown factors appear in various equations, but do not need to be in all of them. What these equations do is to relate all the unknown factors amongt themselves. For example,
$$\left\{
\begin{eqnarray}
3x+2y=1 \\
x5y=6
\end{eqnarray}
\right. $$
is an equation system
with two equations with two unknown
factors (x and y).
Solving this type of problems (a system)
consists in finding a value for each unknown
factor in a way that it applies to all the
equations in the
system. The solution of the example system is:
$$
\begin{eqnarray}
x=1 \\
y=1
\end{eqnarray}$$
There is not always a solution and even there could be an infinite number of solutions. If there is only one solution (one value for each unknown factor, like in the previous example), the system is said to be a consistent dependent system. We will not speak about other kinds of systems.
To solve consistent dependent a system, we need at least the same number of equations as unknown factors. In this section we will resolve linear systems of two equations and two unknown factors with the methods we describe next, which are based on obtaining a first degree equation (a linear equation).
The methods

Substitution (elimination of variables): It consists in isolating
one of the unknown factors (for example x) and
substitute that expression in the other equation.
Thus a first degree equation with the unknown factor y is obtained. Once resolved, we obtain the value
of x using the value of y we know.

Elimination: It consists in operating
the equations, for example, adding or subtracting both
equations so one of the unknown factors disappears.
Thus an equation with only one known factor is obtained.
Equalization: It consists in isolating from
both equations the same unknown factor to be
able to equal both expressions, obtaining one equation with
one unknown factor.
Let us not forget that if we multiply an equation by a number different from 0,
the initial equation and the obtained one
are equivalent. This means that both equations have the same solutions and thereby we can work with one or the other. We will use this property often in the elimination method.
Resolved Systems of Linear Equations
System 1
Show solution
Substitution
We isolate the x from the first equation:
And we substitute it in the second:
Knowing y, we obtain x:
Therefore the solution to the system is
Equalization
We isolate y from both equations:
We equal both expressions and we resolve the equation:
Substituting in the first of the previous equations,
Therefore the solution to the system is
Elimination
We multiply by –2 the first equation,
we add the equations and resolve the
equation we have obtained
We replace the value of y in the first equation and we solve it:
Then, the solution to the system is
System 2
Show solution
Substitution
We isolate the y from the second equation:
And we substitute it in the first equation:
Knowing x, we obtain y:
Therefore the solution to the system is
Equalization
We isolate y from both equations:
We equal both expressions and we resolve the equation:
Substituting in the first of the previous equations,
Therefore the solution to the system is
Elimination
We multiply by 3 the first equation and by 5 the second one:
We change the sign of the second equation (we multiply it by –1) and we add the equations:
We replace the value of y in the first equation and we solve it:
Then, the solution to the system is
System 3
Show solution
Substitution
We isolate y from the first equation:
And we substitute y in the second equation and we resolve it:
Knowing x, we obtain y:
Then, the solution to the system is
Equalization
We isolate y from both equations:
We equal both expressions and we resolve the equation:
We replace the value of x in the first equation:
Hence the solution to the system is
Elimination
We multiply the first equation by 1/5 and by 1/7 the second one:
Thus we avoid the high coefficients that complicate the operations.
We multiply the first equation by 3, the second by 2 and we add them:
We substitute the value of y in the first equation and we resolve it:
Then, the solution to the system is
System 4
Show solution
Substitution
We start by multiplying the first eqaution by the least common multiple of the denominators (thus we avoid fractions):
We substitute the x in the first equation and we resolve it. First,
we multiply the equation by the least common multiple of the denominators:
Knowing y, we obtain x:
Hence the solution to the system is
Equalization
We multiply each equation by the least common multiple of their denominators:
We isolate the x from both equations:
We equal both expressions and we resolve the equation:
We substitute the value of y in the first equation:
Then, the solution to the system is
Elimination
We multiply each equation by the least common multiple of their denominators:
We multiply the first equation by 1 and we add both equations:
We substitute the value of y in the first equation and we resolve it:
Hence the solution to the system is
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