Mister Exam
3x-y=1; 12x=4y+4
The solution
Detail solution
Given the system of equations
$$3 x - y = 1$$
$$12 x = 4 y + 4$$
Let's express from equation 1 x
$$3 x - y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = y + 1$$
$$3 x = y + 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{y + 1}{3}$$
$$x = \frac{y}{3} + \frac{1}{3}$$
Let's try the obtained element x to 2-th equation
$$12 x = 4 y + 4$$
We get:
$$12 \left(\frac{y}{3} + \frac{1}{3}\right) = 4 y + 4$$
so
is always executed
$$3 x - y = 1$$
$$12 x = 4 y + 4$$
Let's express from equation 1 x
$$3 x - y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = y + 1$$
$$3 x = y + 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{y + 1}{3}$$
$$x = \frac{y}{3} + \frac{1}{3}$$
Let's try the obtained element x to 2-th equation
$$12 x = 4 y + 4$$
We get:
$$12 \left(\frac{y}{3} + \frac{1}{3}\right) = 4 y + 4$$
so
is always executed
Rapid solution
$$x_{1} = \frac{y}{3} + \frac{1}{3}$$
=
$$\frac{y}{3} + \frac{1}{3}$$
=
=
$$\frac{y}{3} + \frac{1}{3}$$
=
0.333333333333333 + 0.333333333333333*y
Gaussian elimination
Given the system of equations
$$3 x - y = 1$$
$$12 x = 4 y + 4$$
We give the system of equations to the canonical form
$$3 x - y = 1$$
$$12 x - 4 y = 4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -1 & 1\\12 & -4 & 4\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\12\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}12 - 3 \cdot 4 & -4 - - 4 & \left(-1\right) 4 + 4\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -1 & 1\\0 & 0 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - x_{2} - 1 = 0$$
$$0 - 0 = 0$$
We get the answer:
$$x_{1} = \frac{x_{2}}{3} + \frac{1}{3}$$
where x2 - the free variables
$$3 x - y = 1$$
$$12 x = 4 y + 4$$
We give the system of equations to the canonical form
$$3 x - y = 1$$
$$12 x - 4 y = 4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -1 & 1\\12 & -4 & 4\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\12\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}12 - 3 \cdot 4 & -4 - - 4 & \left(-1\right) 4 + 4\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -1 & 1\\0 & 0 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - x_{2} - 1 = 0$$
$$0 - 0 = 0$$
We get the answer:
$$x_{1} = \frac{x_{2}}{3} + \frac{1}{3}$$
where x2 - the free variables