Mister Exam
5x+4y=23; 2x+6y=18
The solution
You have entered
[src]
5*x + 4*y = 23
$$5 x + 4 y = 23$$
2*x + 6*y = 18
$$2 x + 6 y = 18$$
2*x + 6*y = 18
Detail solution
Given the system of equations
$$5 x + 4 y = 23$$
$$2 x + 6 y = 18$$
Let's express from equation 1 x
$$5 x + 4 y = 23$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 23 - 4 y$$
$$5 x = 23 - 4 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{23 - 4 y}{5}$$
$$x = \frac{23}{5} - \frac{4 y}{5}$$
Let's try the obtained element x to 2-th equation
$$2 x + 6 y = 18$$
We get:
$$6 y + 2 \left(\frac{23}{5} - \frac{4 y}{5}\right) = 18$$
$$\frac{22 y}{5} + \frac{46}{5} = 18$$
We move the free summand 46/5 from the left part to the right part performing the sign change
$$\frac{22 y}{5} = - \frac{46}{5} + 18$$
$$\frac{22 y}{5} = \frac{44}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{22}{5} y}{\frac{22}{5}} = \frac{44}{\frac{22}{5} \cdot 5}$$
$$y = 2$$
Because
$$x = \frac{23}{5} - \frac{4 y}{5}$$
then
$$x = \frac{23}{5} - \frac{8}{5}$$
$$x = 3$$
The answer:
$$x = 3$$
$$y = 2$$
$$5 x + 4 y = 23$$
$$2 x + 6 y = 18$$
Let's express from equation 1 x
$$5 x + 4 y = 23$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 23 - 4 y$$
$$5 x = 23 - 4 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{23 - 4 y}{5}$$
$$x = \frac{23}{5} - \frac{4 y}{5}$$
Let's try the obtained element x to 2-th equation
$$2 x + 6 y = 18$$
We get:
$$6 y + 2 \left(\frac{23}{5} - \frac{4 y}{5}\right) = 18$$
$$\frac{22 y}{5} + \frac{46}{5} = 18$$
We move the free summand 46/5 from the left part to the right part performing the sign change
$$\frac{22 y}{5} = - \frac{46}{5} + 18$$
$$\frac{22 y}{5} = \frac{44}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{22}{5} y}{\frac{22}{5}} = \frac{44}{\frac{22}{5} \cdot 5}$$
$$y = 2$$
Because
$$x = \frac{23}{5} - \frac{4 y}{5}$$
then
$$x = \frac{23}{5} - \frac{8}{5}$$
$$x = 3$$
The answer:
$$x = 3$$
$$y = 2$$
Rapid solution
$$x_{1} = 3$$
=
$$3$$
=
$$y_{1} = 2$$
=
$$2$$
=
=
$$3$$
=
3
$$y_{1} = 2$$
=
$$2$$
=
2
Cramer's rule
$$5 x + 4 y = 23$$
$$2 x + 6 y = 18$$
We give the system of equations to the canonical form
$$5 x + 4 y = 23$$
$$2 x + 6 y = 18$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + 4 x_{2}\\2 x_{1} + 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}23\\18\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 4\\2 & 6\end{matrix}\right] \right)} = 22$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}23 & 4\\18 & 6\end{matrix}\right] \right)}}{22} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 23\\2 & 18\end{matrix}\right] \right)}}{22} = 2$$
$$2 x + 6 y = 18$$
We give the system of equations to the canonical form
$$5 x + 4 y = 23$$
$$2 x + 6 y = 18$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + 4 x_{2}\\2 x_{1} + 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}23\\18\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 4\\2 & 6\end{matrix}\right] \right)} = 22$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}23 & 4\\18 & 6\end{matrix}\right] \right)}}{22} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 23\\2 & 18\end{matrix}\right] \right)}}{22} = 2$$
Gaussian elimination
Given the system of equations
$$5 x + 4 y = 23$$
$$2 x + 6 y = 18$$
We give the system of equations to the canonical form
$$5 x + 4 y = 23$$
$$2 x + 6 y = 18$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 4 & 23\\2 & 6 & 18\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 4 & 23\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 5}{5} & 6 - \frac{2 \cdot 4}{5} & 18 - \frac{2 \cdot 23}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{22}{5} & \frac{44}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 4 & 23\\0 & \frac{22}{5} & \frac{44}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}4\\\frac{22}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{22}{5} & \frac{44}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{0 \cdot 10}{11} & 4 - \frac{10 \cdot 22}{5 \cdot 11} & 23 - \frac{10 \cdot 44}{5 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 15\\0 & \frac{22}{5} & \frac{44}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 15 = 0$$
$$\frac{22 x_{2}}{5} - \frac{44}{5} = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = 2$$
$$5 x + 4 y = 23$$
$$2 x + 6 y = 18$$
We give the system of equations to the canonical form
$$5 x + 4 y = 23$$
$$2 x + 6 y = 18$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 4 & 23\\2 & 6 & 18\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 4 & 23\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 5}{5} & 6 - \frac{2 \cdot 4}{5} & 18 - \frac{2 \cdot 23}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{22}{5} & \frac{44}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 4 & 23\\0 & \frac{22}{5} & \frac{44}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}4\\\frac{22}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{22}{5} & \frac{44}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{0 \cdot 10}{11} & 4 - \frac{10 \cdot 22}{5 \cdot 11} & 23 - \frac{10 \cdot 44}{5 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 15\\0 & \frac{22}{5} & \frac{44}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 15 = 0$$
$$\frac{22 x_{2}}{5} - \frac{44}{5} = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = 2$$