Mister Exam
3x+5y=1; 2x+3y=5
The solution
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3*x + 5*y = 1
$$3 x + 5 y = 1$$
2*x + 3*y = 5
$$2 x + 3 y = 5$$
2*x + 3*y = 5
Detail solution
Given the system of equations
$$3 x + 5 y = 1$$
$$2 x + 3 y = 5$$
Let's express from equation 1 x
$$3 x + 5 y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 1 - 5 y$$
$$3 x = 1 - 5 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{1 - 5 y}{3}$$
$$x = \frac{1}{3} - \frac{5 y}{3}$$
Let's try the obtained element x to 2-th equation
$$2 x + 3 y = 5$$
We get:
$$3 y + 2 \left(\frac{1}{3} - \frac{5 y}{3}\right) = 5$$
$$\frac{2}{3} - \frac{y}{3} = 5$$
We move the free summand 2/3 from the left part to the right part performing the sign change
$$- \frac{y}{3} = - \frac{2}{3} + 5$$
$$- \frac{y}{3} = \frac{13}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{1}{3} y}{- \frac{1}{3}} = \frac{13}{\left(- \frac{1}{3}\right) 3}$$
$$y = -13$$
Because
$$x = \frac{1}{3} - \frac{5 y}{3}$$
then
$$x = \frac{1}{3} - - \frac{65}{3}$$
$$x = 22$$
The answer:
$$x = 22$$
$$y = -13$$
$$3 x + 5 y = 1$$
$$2 x + 3 y = 5$$
Let's express from equation 1 x
$$3 x + 5 y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 1 - 5 y$$
$$3 x = 1 - 5 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{1 - 5 y}{3}$$
$$x = \frac{1}{3} - \frac{5 y}{3}$$
Let's try the obtained element x to 2-th equation
$$2 x + 3 y = 5$$
We get:
$$3 y + 2 \left(\frac{1}{3} - \frac{5 y}{3}\right) = 5$$
$$\frac{2}{3} - \frac{y}{3} = 5$$
We move the free summand 2/3 from the left part to the right part performing the sign change
$$- \frac{y}{3} = - \frac{2}{3} + 5$$
$$- \frac{y}{3} = \frac{13}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{1}{3} y}{- \frac{1}{3}} = \frac{13}{\left(- \frac{1}{3}\right) 3}$$
$$y = -13$$
Because
$$x = \frac{1}{3} - \frac{5 y}{3}$$
then
$$x = \frac{1}{3} - - \frac{65}{3}$$
$$x = 22$$
The answer:
$$x = 22$$
$$y = -13$$
Rapid solution
$$x_{1} = 22$$
=
$$22$$
=
$$y_{1} = -13$$
=
$$-13$$
=
=
$$22$$
=
22
$$y_{1} = -13$$
=
$$-13$$
=
-13
Gaussian elimination
Given the system of equations
$$3 x + 5 y = 1$$
$$2 x + 3 y = 5$$
We give the system of equations to the canonical form
$$3 x + 5 y = 1$$
$$2 x + 3 y = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 5 & 1\\2 & 3 & 5\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 5 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 3}{3} & 3 - \frac{2 \cdot 5}{3} & \frac{\left(-1\right) 2}{3} + 5\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{1}{3} & \frac{13}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 5 & 1\\0 & - \frac{1}{3} & \frac{13}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}5\\- \frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{1}{3} & \frac{13}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \left(-15\right) 0 & 5 - - -5 & 1 - - 65\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 66\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 66\\0 & - \frac{1}{3} & \frac{13}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 66 = 0$$
$$- \frac{x_{2}}{3} - \frac{13}{3} = 0$$
We get the answer:
$$x_{1} = 22$$
$$x_{2} = -13$$
$$3 x + 5 y = 1$$
$$2 x + 3 y = 5$$
We give the system of equations to the canonical form
$$3 x + 5 y = 1$$
$$2 x + 3 y = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 5 & 1\\2 & 3 & 5\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 5 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 3}{3} & 3 - \frac{2 \cdot 5}{3} & \frac{\left(-1\right) 2}{3} + 5\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{1}{3} & \frac{13}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 5 & 1\\0 & - \frac{1}{3} & \frac{13}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}5\\- \frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{1}{3} & \frac{13}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \left(-15\right) 0 & 5 - - -5 & 1 - - 65\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 66\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 66\\0 & - \frac{1}{3} & \frac{13}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 66 = 0$$
$$- \frac{x_{2}}{3} - \frac{13}{3} = 0$$
We get the answer:
$$x_{1} = 22$$
$$x_{2} = -13$$
Cramer's rule
$$3 x + 5 y = 1$$
$$2 x + 3 y = 5$$
We give the system of equations to the canonical form
$$3 x + 5 y = 1$$
$$2 x + 3 y = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + 5 x_{2}\\2 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\5\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 5\\2 & 3\end{matrix}\right] \right)} = -1$$
, 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 )
$$x_{1} = - \operatorname{det}{\left(\left[\begin{matrix}1 & 5\\5 & 3\end{matrix}\right] \right)} = 22$$
$$x_{2} = - \operatorname{det}{\left(\left[\begin{matrix}3 & 1\\2 & 5\end{matrix}\right] \right)} = -13$$
$$2 x + 3 y = 5$$
We give the system of equations to the canonical form
$$3 x + 5 y = 1$$
$$2 x + 3 y = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + 5 x_{2}\\2 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\5\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 5\\2 & 3\end{matrix}\right] \right)} = -1$$
, 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 )
$$x_{1} = - \operatorname{det}{\left(\left[\begin{matrix}1 & 5\\5 & 3\end{matrix}\right] \right)} = 22$$
$$x_{2} = - \operatorname{det}{\left(\left[\begin{matrix}3 & 1\\2 & 5\end{matrix}\right] \right)} = -13$$