Mister Exam
4х+2у=5; 4х-6у=-7
The solution
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4*x + 2*y = 5
$$4 x + 2 y = 5$$
4*x - 6*y = -7
$$4 x - 6 y = -7$$
4*x - 6*y = -7
Detail solution
Given the system of equations
$$4 x + 2 y = 5$$
$$4 x - 6 y = -7$$
Let's express from equation 1 x
$$4 x + 2 y = 5$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 5 - 2 y$$
$$4 x = 5 - 2 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{5 - 2 y}{4}$$
$$x = \frac{5}{4} - \frac{y}{2}$$
Let's try the obtained element x to 2-th equation
$$4 x - 6 y = -7$$
We get:
$$- 6 y + 4 \left(\frac{5}{4} - \frac{y}{2}\right) = -7$$
$$5 - 8 y = -7$$
We move the free summand 5 from the left part to the right part performing the sign change
$$- 8 y = -7 - 5$$
$$- 8 y = -12$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 8 y}{-8} = - \frac{12}{-8}$$
$$y = \frac{3}{2}$$
Because
$$x = \frac{5}{4} - \frac{y}{2}$$
then
$$x = \frac{5}{4} - \frac{3}{4}$$
$$x = \frac{1}{2}$$
The answer:
$$x = \frac{1}{2}$$
$$y = \frac{3}{2}$$
$$4 x + 2 y = 5$$
$$4 x - 6 y = -7$$
Let's express from equation 1 x
$$4 x + 2 y = 5$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 5 - 2 y$$
$$4 x = 5 - 2 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{5 - 2 y}{4}$$
$$x = \frac{5}{4} - \frac{y}{2}$$
Let's try the obtained element x to 2-th equation
$$4 x - 6 y = -7$$
We get:
$$- 6 y + 4 \left(\frac{5}{4} - \frac{y}{2}\right) = -7$$
$$5 - 8 y = -7$$
We move the free summand 5 from the left part to the right part performing the sign change
$$- 8 y = -7 - 5$$
$$- 8 y = -12$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 8 y}{-8} = - \frac{12}{-8}$$
$$y = \frac{3}{2}$$
Because
$$x = \frac{5}{4} - \frac{y}{2}$$
then
$$x = \frac{5}{4} - \frac{3}{4}$$
$$x = \frac{1}{2}$$
The answer:
$$x = \frac{1}{2}$$
$$y = \frac{3}{2}$$
Rapid solution
$$x_{1} = \frac{1}{2}$$
=
$$\frac{1}{2}$$
=
$$y_{1} = \frac{3}{2}$$
=
$$\frac{3}{2}$$
=
=
$$\frac{1}{2}$$
=
0.5
$$y_{1} = \frac{3}{2}$$
=
$$\frac{3}{2}$$
=
1.5
Gaussian elimination
Given the system of equations
$$4 x + 2 y = 5$$
$$4 x - 6 y = -7$$
We give the system of equations to the canonical form
$$4 x + 2 y = 5$$
$$4 x - 6 y = -7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 2 & 5\\4 & -6 & -7\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 2 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 4 + 4 & -6 + \left(-1\right) 2 & -7 + \left(-1\right) 5\end{matrix}\right] = \left[\begin{matrix}0 & -8 & -12\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 2 & 5\\0 & -8 & -12\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}2\\-8\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -8 & -12\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{\left(-1\right) 0}{4} & 2 - - -2 & 5 - - -3\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 2\\0 & -8 & -12\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 2 = 0$$
$$12 - 8 x_{2} = 0$$
We get the answer:
$$x_{1} = \frac{1}{2}$$
$$x_{2} = \frac{3}{2}$$
$$4 x + 2 y = 5$$
$$4 x - 6 y = -7$$
We give the system of equations to the canonical form
$$4 x + 2 y = 5$$
$$4 x - 6 y = -7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 2 & 5\\4 & -6 & -7\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 2 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 4 + 4 & -6 + \left(-1\right) 2 & -7 + \left(-1\right) 5\end{matrix}\right] = \left[\begin{matrix}0 & -8 & -12\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 2 & 5\\0 & -8 & -12\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}2\\-8\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -8 & -12\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{\left(-1\right) 0}{4} & 2 - - -2 & 5 - - -3\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 2\\0 & -8 & -12\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 2 = 0$$
$$12 - 8 x_{2} = 0$$
We get the answer:
$$x_{1} = \frac{1}{2}$$
$$x_{2} = \frac{3}{2}$$
Cramer's rule
$$4 x + 2 y = 5$$
$$4 x - 6 y = -7$$
We give the system of equations to the canonical form
$$4 x + 2 y = 5$$
$$4 x - 6 y = -7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + 2 x_{2}\\4 x_{1} - 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}5\\-7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 2\\4 & -6\end{matrix}\right] \right)} = -32$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 2\\-7 & -6\end{matrix}\right] \right)}}{32} = \frac{1}{2}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 5\\4 & -7\end{matrix}\right] \right)}}{32} = \frac{3}{2}$$
$$4 x - 6 y = -7$$
We give the system of equations to the canonical form
$$4 x + 2 y = 5$$
$$4 x - 6 y = -7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + 2 x_{2}\\4 x_{1} - 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}5\\-7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 2\\4 & -6\end{matrix}\right] \right)} = -32$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 2\\-7 & -6\end{matrix}\right] \right)}}{32} = \frac{1}{2}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 5\\4 & -7\end{matrix}\right] \right)}}{32} = \frac{3}{2}$$