Mister Exam
4x+12y=1; 6x+9y=1
The solution
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4*x + 12*y = 1
$$4 x + 12 y = 1$$
6*x + 9*y = 1
$$6 x + 9 y = 1$$
6*x + 9*y = 1
Detail solution
Given the system of equations
$$4 x + 12 y = 1$$
$$6 x + 9 y = 1$$
Let's express from equation 1 x
$$4 x + 12 y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 1 - 12 y$$
$$4 x = 1 - 12 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{1 - 12 y}{4}$$
$$x = \frac{1}{4} - 3 y$$
Let's try the obtained element x to 2-th equation
$$6 x + 9 y = 1$$
We get:
$$9 y + 6 \left(\frac{1}{4} - 3 y\right) = 1$$
$$\frac{3}{2} - 9 y = 1$$
We move the free summand 3/2 from the left part to the right part performing the sign change
$$- 9 y = - \frac{3}{2} + 1$$
$$- 9 y = - \frac{1}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 9 y}{-9} = - \frac{1}{\left(-9\right) 2}$$
$$y = \frac{1}{18}$$
Because
$$x = \frac{1}{4} - 3 y$$
then
$$x = \frac{1}{4} - \frac{1}{6}$$
$$x = \frac{1}{12}$$
The answer:
$$x = \frac{1}{12}$$
$$y = \frac{1}{18}$$
$$4 x + 12 y = 1$$
$$6 x + 9 y = 1$$
Let's express from equation 1 x
$$4 x + 12 y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 1 - 12 y$$
$$4 x = 1 - 12 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{1 - 12 y}{4}$$
$$x = \frac{1}{4} - 3 y$$
Let's try the obtained element x to 2-th equation
$$6 x + 9 y = 1$$
We get:
$$9 y + 6 \left(\frac{1}{4} - 3 y\right) = 1$$
$$\frac{3}{2} - 9 y = 1$$
We move the free summand 3/2 from the left part to the right part performing the sign change
$$- 9 y = - \frac{3}{2} + 1$$
$$- 9 y = - \frac{1}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 9 y}{-9} = - \frac{1}{\left(-9\right) 2}$$
$$y = \frac{1}{18}$$
Because
$$x = \frac{1}{4} - 3 y$$
then
$$x = \frac{1}{4} - \frac{1}{6}$$
$$x = \frac{1}{12}$$
The answer:
$$x = \frac{1}{12}$$
$$y = \frac{1}{18}$$
Rapid solution
$$x_{1} = \frac{1}{12}$$
=
$$\frac{1}{12}$$
=
$$y_{1} = \frac{1}{18}$$
=
$$\frac{1}{18}$$
=
=
$$\frac{1}{12}$$
=
0.0833333333333333
$$y_{1} = \frac{1}{18}$$
=
$$\frac{1}{18}$$
=
0.0555555555555556
Cramer's rule
$$4 x + 12 y = 1$$
$$6 x + 9 y = 1$$
We give the system of equations to the canonical form
$$4 x + 12 y = 1$$
$$6 x + 9 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + 12 x_{2}\\6 x_{1} + 9 x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\1\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 12\\6 & 9\end{matrix}\right] \right)} = -36$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 12\\1 & 9\end{matrix}\right] \right)}}{36} = \frac{1}{12}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 1\\6 & 1\end{matrix}\right] \right)}}{36} = \frac{1}{18}$$
$$6 x + 9 y = 1$$
We give the system of equations to the canonical form
$$4 x + 12 y = 1$$
$$6 x + 9 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + 12 x_{2}\\6 x_{1} + 9 x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\1\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 12\\6 & 9\end{matrix}\right] \right)} = -36$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 12\\1 & 9\end{matrix}\right] \right)}}{36} = \frac{1}{12}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 1\\6 & 1\end{matrix}\right] \right)}}{36} = \frac{1}{18}$$
Gaussian elimination
Given the system of equations
$$4 x + 12 y = 1$$
$$6 x + 9 y = 1$$
We give the system of equations to the canonical form
$$4 x + 12 y = 1$$
$$6 x + 9 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 12 & 1\\6 & 9 & 1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 12 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{3 \cdot 4}{2} & 9 - \frac{3 \cdot 12}{2} & \frac{\left(-1\right) 3}{2} + 1\end{matrix}\right] = \left[\begin{matrix}0 & -9 & - \frac{1}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 12 & 1\\0 & -9 & - \frac{1}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}12\\-9\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -9 & - \frac{1}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{\left(-4\right) 0}{3} & 12 - - -12 & 1 - - \frac{-2}{3}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & \frac{1}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & \frac{1}{3}\\0 & -9 & - \frac{1}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - \frac{1}{3} = 0$$
$$\frac{1}{2} - 9 x_{2} = 0$$
We get the answer:
$$x_{1} = \frac{1}{12}$$
$$x_{2} = \frac{1}{18}$$
$$4 x + 12 y = 1$$
$$6 x + 9 y = 1$$
We give the system of equations to the canonical form
$$4 x + 12 y = 1$$
$$6 x + 9 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 12 & 1\\6 & 9 & 1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 12 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{3 \cdot 4}{2} & 9 - \frac{3 \cdot 12}{2} & \frac{\left(-1\right) 3}{2} + 1\end{matrix}\right] = \left[\begin{matrix}0 & -9 & - \frac{1}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 12 & 1\\0 & -9 & - \frac{1}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}12\\-9\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -9 & - \frac{1}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{\left(-4\right) 0}{3} & 12 - - -12 & 1 - - \frac{-2}{3}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & \frac{1}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & \frac{1}{3}\\0 & -9 & - \frac{1}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - \frac{1}{3} = 0$$
$$\frac{1}{2} - 9 x_{2} = 0$$
We get the answer:
$$x_{1} = \frac{1}{12}$$
$$x_{2} = \frac{1}{18}$$
Numerical answer
[src]
x1 = 0.08333333333333333 y1 = 0.05555555555555556
x1 = 0.08333333333333333 y1 = 0.05555555555555556