Mister Exam
5x-6y=1; 3x+4y=12
The solution
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5*x - 6*y = 1
$$5 x - 6 y = 1$$
3*x + 4*y = 12
$$3 x + 4 y = 12$$
3*x + 4*y = 12
Detail solution
Given the system of equations
$$5 x - 6 y = 1$$
$$3 x + 4 y = 12$$
Let's express from equation 1 x
$$5 x - 6 y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 6 y + 1$$
$$5 x = 6 y + 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{6 y + 1}{5}$$
$$x = \frac{6 y}{5} + \frac{1}{5}$$
Let's try the obtained element x to 2-th equation
$$3 x + 4 y = 12$$
We get:
$$4 y + 3 \left(\frac{6 y}{5} + \frac{1}{5}\right) = 12$$
$$\frac{38 y}{5} + \frac{3}{5} = 12$$
We move the free summand 3/5 from the left part to the right part performing the sign change
$$\frac{38 y}{5} = - \frac{3}{5} + 12$$
$$\frac{38 y}{5} = \frac{57}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{38}{5} y}{\frac{38}{5}} = \frac{57}{5 \frac{38}{5}}$$
$$y = \frac{3}{2}$$
Because
$$x = \frac{6 y}{5} + \frac{1}{5}$$
then
$$x = \frac{1}{5} + \frac{3 \cdot 6}{2 \cdot 5}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = \frac{3}{2}$$
$$5 x - 6 y = 1$$
$$3 x + 4 y = 12$$
Let's express from equation 1 x
$$5 x - 6 y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 6 y + 1$$
$$5 x = 6 y + 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{6 y + 1}{5}$$
$$x = \frac{6 y}{5} + \frac{1}{5}$$
Let's try the obtained element x to 2-th equation
$$3 x + 4 y = 12$$
We get:
$$4 y + 3 \left(\frac{6 y}{5} + \frac{1}{5}\right) = 12$$
$$\frac{38 y}{5} + \frac{3}{5} = 12$$
We move the free summand 3/5 from the left part to the right part performing the sign change
$$\frac{38 y}{5} = - \frac{3}{5} + 12$$
$$\frac{38 y}{5} = \frac{57}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{38}{5} y}{\frac{38}{5}} = \frac{57}{5 \frac{38}{5}}$$
$$y = \frac{3}{2}$$
Because
$$x = \frac{6 y}{5} + \frac{1}{5}$$
then
$$x = \frac{1}{5} + \frac{3 \cdot 6}{2 \cdot 5}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = \frac{3}{2}$$
Rapid solution
$$x_{1} = 2$$
=
$$2$$
=
$$y_{1} = \frac{3}{2}$$
=
$$\frac{3}{2}$$
=
=
$$2$$
=
2
$$y_{1} = \frac{3}{2}$$
=
$$\frac{3}{2}$$
=
1.5
Cramer's rule
$$5 x - 6 y = 1$$
$$3 x + 4 y = 12$$
We give the system of equations to the canonical form
$$5 x - 6 y = 1$$
$$3 x + 4 y = 12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 6 x_{2}\\3 x_{1} + 4 x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\12\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -6\\3 & 4\end{matrix}\right] \right)} = 38$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -6\\12 & 4\end{matrix}\right] \right)}}{38} = 2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 1\\3 & 12\end{matrix}\right] \right)}}{38} = \frac{3}{2}$$
$$3 x + 4 y = 12$$
We give the system of equations to the canonical form
$$5 x - 6 y = 1$$
$$3 x + 4 y = 12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 6 x_{2}\\3 x_{1} + 4 x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\12\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -6\\3 & 4\end{matrix}\right] \right)} = 38$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -6\\12 & 4\end{matrix}\right] \right)}}{38} = 2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 1\\3 & 12\end{matrix}\right] \right)}}{38} = \frac{3}{2}$$
Gaussian elimination
Given the system of equations
$$5 x - 6 y = 1$$
$$3 x + 4 y = 12$$
We give the system of equations to the canonical form
$$5 x - 6 y = 1$$
$$3 x + 4 y = 12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -6 & 1\\3 & 4 & 12\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -6 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{3 \cdot 5}{5} & 4 - - \frac{18}{5} & \frac{\left(-1\right) 3}{5} + 12\end{matrix}\right] = \left[\begin{matrix}0 & \frac{38}{5} & \frac{57}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -6 & 1\\0 & \frac{38}{5} & \frac{57}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-6\\\frac{38}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{38}{5} & \frac{57}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-15\right) 0}{19} & -6 - \frac{\left(-15\right) 38}{5 \cdot 19} & 1 - \frac{\left(-15\right) 57}{5 \cdot 19}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 10\\0 & \frac{38}{5} & \frac{57}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 10 = 0$$
$$\frac{38 x_{2}}{5} - \frac{57}{5} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = \frac{3}{2}$$
$$5 x - 6 y = 1$$
$$3 x + 4 y = 12$$
We give the system of equations to the canonical form
$$5 x - 6 y = 1$$
$$3 x + 4 y = 12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -6 & 1\\3 & 4 & 12\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -6 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{3 \cdot 5}{5} & 4 - - \frac{18}{5} & \frac{\left(-1\right) 3}{5} + 12\end{matrix}\right] = \left[\begin{matrix}0 & \frac{38}{5} & \frac{57}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -6 & 1\\0 & \frac{38}{5} & \frac{57}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-6\\\frac{38}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{38}{5} & \frac{57}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-15\right) 0}{19} & -6 - \frac{\left(-15\right) 38}{5 \cdot 19} & 1 - \frac{\left(-15\right) 57}{5 \cdot 19}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 10\\0 & \frac{38}{5} & \frac{57}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 10 = 0$$
$$\frac{38 x_{2}}{5} - \frac{57}{5} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = \frac{3}{2}$$