Mister Exam
4х-5у=12; 6х+11у=-19
The solution
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4*x - 5*y = 12
$$4 x - 5 y = 12$$
6*x + 11*y = -19
$$6 x + 11 y = -19$$
6*x + 11*y = -19
Detail solution
Given the system of equations
$$4 x - 5 y = 12$$
$$6 x + 11 y = -19$$
Let's express from equation 1 x
$$4 x - 5 y = 12$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 5 y + 12$$
$$4 x = 5 y + 12$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{5 y + 12}{4}$$
$$x = \frac{5 y}{4} + 3$$
Let's try the obtained element x to 2-th equation
$$6 x + 11 y = -19$$
We get:
$$11 y + 6 \left(\frac{5 y}{4} + 3\right) = -19$$
$$\frac{37 y}{2} + 18 = -19$$
We move the free summand 18 from the left part to the right part performing the sign change
$$\frac{37 y}{2} = -19 - 18$$
$$\frac{37 y}{2} = -37$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{37}{2} y}{\frac{37}{2}} = - \frac{37}{\frac{37}{2}}$$
$$y = -2$$
Because
$$x = \frac{5 y}{4} + 3$$
then
$$x = \frac{\left(-2\right) 5}{4} + 3$$
$$x = \frac{1}{2}$$
The answer:
$$x = \frac{1}{2}$$
$$y = -2$$
$$4 x - 5 y = 12$$
$$6 x + 11 y = -19$$
Let's express from equation 1 x
$$4 x - 5 y = 12$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 5 y + 12$$
$$4 x = 5 y + 12$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{5 y + 12}{4}$$
$$x = \frac{5 y}{4} + 3$$
Let's try the obtained element x to 2-th equation
$$6 x + 11 y = -19$$
We get:
$$11 y + 6 \left(\frac{5 y}{4} + 3\right) = -19$$
$$\frac{37 y}{2} + 18 = -19$$
We move the free summand 18 from the left part to the right part performing the sign change
$$\frac{37 y}{2} = -19 - 18$$
$$\frac{37 y}{2} = -37$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{37}{2} y}{\frac{37}{2}} = - \frac{37}{\frac{37}{2}}$$
$$y = -2$$
Because
$$x = \frac{5 y}{4} + 3$$
then
$$x = \frac{\left(-2\right) 5}{4} + 3$$
$$x = \frac{1}{2}$$
The answer:
$$x = \frac{1}{2}$$
$$y = -2$$
Rapid solution
$$x_{1} = \frac{1}{2}$$
=
$$\frac{1}{2}$$
=
$$y_{1} = -2$$
=
$$-2$$
=
=
$$\frac{1}{2}$$
=
0.5
$$y_{1} = -2$$
=
$$-2$$
=
-2
Cramer's rule
$$4 x - 5 y = 12$$
$$6 x + 11 y = -19$$
We give the system of equations to the canonical form
$$4 x - 5 y = 12$$
$$6 x + 11 y = -19$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} - 5 x_{2}\\6 x_{1} + 11 x_{2}\end{matrix}\right] = \left[\begin{matrix}12\\-19\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & -5\\6 & 11\end{matrix}\right] \right)} = 74$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}12 & -5\\-19 & 11\end{matrix}\right] \right)}}{74} = \frac{1}{2}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 12\\6 & -19\end{matrix}\right] \right)}}{74} = -2$$
$$6 x + 11 y = -19$$
We give the system of equations to the canonical form
$$4 x - 5 y = 12$$
$$6 x + 11 y = -19$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} - 5 x_{2}\\6 x_{1} + 11 x_{2}\end{matrix}\right] = \left[\begin{matrix}12\\-19\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & -5\\6 & 11\end{matrix}\right] \right)} = 74$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}12 & -5\\-19 & 11\end{matrix}\right] \right)}}{74} = \frac{1}{2}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 12\\6 & -19\end{matrix}\right] \right)}}{74} = -2$$
Gaussian elimination
Given the system of equations
$$4 x - 5 y = 12$$
$$6 x + 11 y = -19$$
We give the system of equations to the canonical form
$$4 x - 5 y = 12$$
$$6 x + 11 y = -19$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & -5 & 12\\6 & 11 & -19\end{matrix}\right]$$
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
$$\left[\begin{matrix}4 & -5 & 12\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{3 \cdot 4}{2} & 11 - - \frac{15}{2} & -19 - \frac{3 \cdot 12}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{37}{2} & -37\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & -5 & 12\\0 & \frac{37}{2} & -37\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{37}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{37}{2} & -37\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{\left(-10\right) 0}{37} & -5 - \frac{\left(-10\right) 37}{2 \cdot 37} & 12 - - -10\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 2\\0 & \frac{37}{2} & -37\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 2 = 0$$
$$\frac{37 x_{2}}{2} + 37 = 0$$
We get the answer:
$$x_{1} = \frac{1}{2}$$
$$x_{2} = -2$$
$$4 x - 5 y = 12$$
$$6 x + 11 y = -19$$
We give the system of equations to the canonical form
$$4 x - 5 y = 12$$
$$6 x + 11 y = -19$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & -5 & 12\\6 & 11 & -19\end{matrix}\right]$$
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
$$\left[\begin{matrix}4 & -5 & 12\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{3 \cdot 4}{2} & 11 - - \frac{15}{2} & -19 - \frac{3 \cdot 12}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{37}{2} & -37\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & -5 & 12\\0 & \frac{37}{2} & -37\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{37}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{37}{2} & -37\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{\left(-10\right) 0}{37} & -5 - \frac{\left(-10\right) 37}{2 \cdot 37} & 12 - - -10\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 2\\0 & \frac{37}{2} & -37\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 2 = 0$$
$$\frac{37 x_{2}}{2} + 37 = 0$$
We get the answer:
$$x_{1} = \frac{1}{2}$$
$$x_{2} = -2$$