Mister Exam
5х+4у=-7; 5х-6у=23
The solution
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5*x + 4*y = -7
$$5 x + 4 y = -7$$
5*x - 6*y = 23
$$5 x - 6 y = 23$$
5*x - 6*y = 23
Detail solution
Given the system of equations
$$5 x + 4 y = -7$$
$$5 x - 6 y = 23$$
Let's express from equation 1 x
$$5 x + 4 y = -7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = - 4 y - 7$$
$$5 x = - 4 y - 7$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{- 4 y - 7}{5}$$
$$x = - \frac{4 y}{5} - \frac{7}{5}$$
Let's try the obtained element x to 2-th equation
$$5 x - 6 y = 23$$
We get:
$$- 6 y + 5 \left(- \frac{4 y}{5} - \frac{7}{5}\right) = 23$$
$$- 10 y - 7 = 23$$
We move the free summand -7 from the left part to the right part performing the sign change
$$- 10 y = 7 + 23$$
$$- 10 y = 30$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 10 y}{-10} = \frac{30}{-10}$$
$$y = -3$$
Because
$$x = - \frac{4 y}{5} - \frac{7}{5}$$
then
$$x = - \frac{7}{5} - - \frac{12}{5}$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = -3$$
$$5 x + 4 y = -7$$
$$5 x - 6 y = 23$$
Let's express from equation 1 x
$$5 x + 4 y = -7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = - 4 y - 7$$
$$5 x = - 4 y - 7$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{- 4 y - 7}{5}$$
$$x = - \frac{4 y}{5} - \frac{7}{5}$$
Let's try the obtained element x to 2-th equation
$$5 x - 6 y = 23$$
We get:
$$- 6 y + 5 \left(- \frac{4 y}{5} - \frac{7}{5}\right) = 23$$
$$- 10 y - 7 = 23$$
We move the free summand -7 from the left part to the right part performing the sign change
$$- 10 y = 7 + 23$$
$$- 10 y = 30$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 10 y}{-10} = \frac{30}{-10}$$
$$y = -3$$
Because
$$x = - \frac{4 y}{5} - \frac{7}{5}$$
then
$$x = - \frac{7}{5} - - \frac{12}{5}$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = -3$$
Rapid solution
$$x_{1} = 1$$
=
$$1$$
=
$$y_{1} = -3$$
=
$$-3$$
=
=
$$1$$
=
1
$$y_{1} = -3$$
=
$$-3$$
=
-3
Cramer's rule
$$5 x + 4 y = -7$$
$$5 x - 6 y = 23$$
We give the system of equations to the canonical form
$$5 x + 4 y = -7$$
$$5 x - 6 y = 23$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + 4 x_{2}\\5 x_{1} - 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}-7\\23\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 4\\5 & -6\end{matrix}\right] \right)} = -50$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-7 & 4\\23 & -6\end{matrix}\right] \right)}}{50} = 1$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -7\\5 & 23\end{matrix}\right] \right)}}{50} = -3$$
$$5 x - 6 y = 23$$
We give the system of equations to the canonical form
$$5 x + 4 y = -7$$
$$5 x - 6 y = 23$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + 4 x_{2}\\5 x_{1} - 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}-7\\23\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 4\\5 & -6\end{matrix}\right] \right)} = -50$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-7 & 4\\23 & -6\end{matrix}\right] \right)}}{50} = 1$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -7\\5 & 23\end{matrix}\right] \right)}}{50} = -3$$
Gaussian elimination
Given the system of equations
$$5 x + 4 y = -7$$
$$5 x - 6 y = 23$$
We give the system of equations to the canonical form
$$5 x + 4 y = -7$$
$$5 x - 6 y = 23$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 4 & -7\\5 & -6 & 23\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 4 & -7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 5 + 5 & -6 + \left(-1\right) 4 & 23 - -7\end{matrix}\right] = \left[\begin{matrix}0 & -10 & 30\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 4 & -7\\0 & -10 & 30\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}4\\-10\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -10 & 30\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-2\right) 0}{5} & 4 - - -4 & -7 - \frac{\left(-2\right) 30}{5}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 5\\0 & -10 & 30\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 5 = 0$$
$$- 10 x_{2} - 30 = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -3$$
$$5 x + 4 y = -7$$
$$5 x - 6 y = 23$$
We give the system of equations to the canonical form
$$5 x + 4 y = -7$$
$$5 x - 6 y = 23$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 4 & -7\\5 & -6 & 23\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 4 & -7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 5 + 5 & -6 + \left(-1\right) 4 & 23 - -7\end{matrix}\right] = \left[\begin{matrix}0 & -10 & 30\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 4 & -7\\0 & -10 & 30\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}4\\-10\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -10 & 30\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-2\right) 0}{5} & 4 - - -4 & -7 - \frac{\left(-2\right) 30}{5}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 5\\0 & -10 & 30\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 5 = 0$$
$$- 10 x_{2} - 30 = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -3$$