Mister Exam
3х+y=7; -5x+2y=3
The solution
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3*x + y = 7
$$3 x + y = 7$$
-5*x + 2*y = 3
$$- 5 x + 2 y = 3$$
-5*x + 2*y = 3
Detail solution
Given the system of equations
$$3 x + y = 7$$
$$- 5 x + 2 y = 3$$
Let's express from equation 1 x
$$3 x + y = 7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 7 - y$$
$$3 x = 7 - y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{7 - y}{3}$$
$$x = \frac{7}{3} - \frac{y}{3}$$
Let's try the obtained element x to 2-th equation
$$- 5 x + 2 y = 3$$
We get:
$$2 y - 5 \left(\frac{7}{3} - \frac{y}{3}\right) = 3$$
$$\frac{11 y}{3} - \frac{35}{3} = 3$$
We move the free summand -35/3 from the left part to the right part performing the sign change
$$\frac{11 y}{3} = 3 + \frac{35}{3}$$
$$\frac{11 y}{3} = \frac{44}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{11}{3} y}{\frac{11}{3}} = \frac{44}{3 \frac{11}{3}}$$
$$y = 4$$
Because
$$x = \frac{7}{3} - \frac{y}{3}$$
then
$$x = \frac{7}{3} - \frac{4}{3}$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = 4$$
$$3 x + y = 7$$
$$- 5 x + 2 y = 3$$
Let's express from equation 1 x
$$3 x + y = 7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 7 - y$$
$$3 x = 7 - y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{7 - y}{3}$$
$$x = \frac{7}{3} - \frac{y}{3}$$
Let's try the obtained element x to 2-th equation
$$- 5 x + 2 y = 3$$
We get:
$$2 y - 5 \left(\frac{7}{3} - \frac{y}{3}\right) = 3$$
$$\frac{11 y}{3} - \frac{35}{3} = 3$$
We move the free summand -35/3 from the left part to the right part performing the sign change
$$\frac{11 y}{3} = 3 + \frac{35}{3}$$
$$\frac{11 y}{3} = \frac{44}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{11}{3} y}{\frac{11}{3}} = \frac{44}{3 \frac{11}{3}}$$
$$y = 4$$
Because
$$x = \frac{7}{3} - \frac{y}{3}$$
then
$$x = \frac{7}{3} - \frac{4}{3}$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = 4$$
Rapid solution
$$x_{1} = 1$$
=
$$1$$
=
$$y_{1} = 4$$
=
$$4$$
=
=
$$1$$
=
1
$$y_{1} = 4$$
=
$$4$$
=
4
Gaussian elimination
Given the system of equations
$$3 x + y = 7$$
$$- 5 x + 2 y = 3$$
We give the system of equations to the canonical form
$$3 x + y = 7$$
$$- 5 x + 2 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 1 & 7\\-5 & 2 & 3\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\-5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 1 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-5 - \frac{\left(-5\right) 3}{3} & 2 - - \frac{5}{3} & 3 - \frac{\left(-5\right) 7}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{11}{3} & \frac{44}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 1 & 7\\0 & \frac{11}{3} & \frac{44}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\\frac{11}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{11}{3} & \frac{44}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{0 \cdot 3}{11} & 1 - \frac{3 \cdot 11}{3 \cdot 11} & 7 - \frac{3 \cdot 44}{3 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 3\\0 & \frac{11}{3} & \frac{44}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 3 = 0$$
$$\frac{11 x_{2}}{3} - \frac{44}{3} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = 4$$
$$3 x + y = 7$$
$$- 5 x + 2 y = 3$$
We give the system of equations to the canonical form
$$3 x + y = 7$$
$$- 5 x + 2 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 1 & 7\\-5 & 2 & 3\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\-5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 1 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-5 - \frac{\left(-5\right) 3}{3} & 2 - - \frac{5}{3} & 3 - \frac{\left(-5\right) 7}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{11}{3} & \frac{44}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 1 & 7\\0 & \frac{11}{3} & \frac{44}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\\frac{11}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{11}{3} & \frac{44}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{0 \cdot 3}{11} & 1 - \frac{3 \cdot 11}{3 \cdot 11} & 7 - \frac{3 \cdot 44}{3 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 3\\0 & \frac{11}{3} & \frac{44}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 3 = 0$$
$$\frac{11 x_{2}}{3} - \frac{44}{3} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = 4$$
Cramer's rule
$$3 x + y = 7$$
$$- 5 x + 2 y = 3$$
We give the system of equations to the canonical form
$$3 x + y = 7$$
$$- 5 x + 2 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + x_{2}\\- 5 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}7\\3\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 1\\-5 & 2\end{matrix}\right] \right)} = 11$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}7 & 1\\3 & 2\end{matrix}\right] \right)}}{11} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 7\\-5 & 3\end{matrix}\right] \right)}}{11} = 4$$
$$- 5 x + 2 y = 3$$
We give the system of equations to the canonical form
$$3 x + y = 7$$
$$- 5 x + 2 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + x_{2}\\- 5 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}7\\3\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 1\\-5 & 2\end{matrix}\right] \right)} = 11$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}7 & 1\\3 & 2\end{matrix}\right] \right)}}{11} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 7\\-5 & 3\end{matrix}\right] \right)}}{11} = 4$$