Mister Exam
3х-5у=-21; 7х+10у=16
The solution
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3*x - 5*y = -21
$$3 x - 5 y = -21$$
7*x + 10*y = 16
$$7 x + 10 y = 16$$
7*x + 10*y = 16
Detail solution
Given the system of equations
$$3 x - 5 y = -21$$
$$7 x + 10 y = 16$$
Let's express from equation 1 x
$$3 x - 5 y = -21$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 5 y - 21$$
$$3 x = 5 y - 21$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{5 y - 21}{3}$$
$$x = \frac{5 y}{3} - 7$$
Let's try the obtained element x to 2-th equation
$$7 x + 10 y = 16$$
We get:
$$10 y + 7 \left(\frac{5 y}{3} - 7\right) = 16$$
$$\frac{65 y}{3} - 49 = 16$$
We move the free summand -49 from the left part to the right part performing the sign change
$$\frac{65 y}{3} = 16 + 49$$
$$\frac{65 y}{3} = 65$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{65}{3} y}{\frac{65}{3}} = \frac{65}{\frac{65}{3}}$$
$$y = 3$$
Because
$$x = \frac{5 y}{3} - 7$$
then
$$x = -7 + \frac{3 \cdot 5}{3}$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = 3$$
$$3 x - 5 y = -21$$
$$7 x + 10 y = 16$$
Let's express from equation 1 x
$$3 x - 5 y = -21$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 5 y - 21$$
$$3 x = 5 y - 21$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{5 y - 21}{3}$$
$$x = \frac{5 y}{3} - 7$$
Let's try the obtained element x to 2-th equation
$$7 x + 10 y = 16$$
We get:
$$10 y + 7 \left(\frac{5 y}{3} - 7\right) = 16$$
$$\frac{65 y}{3} - 49 = 16$$
We move the free summand -49 from the left part to the right part performing the sign change
$$\frac{65 y}{3} = 16 + 49$$
$$\frac{65 y}{3} = 65$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{65}{3} y}{\frac{65}{3}} = \frac{65}{\frac{65}{3}}$$
$$y = 3$$
Because
$$x = \frac{5 y}{3} - 7$$
then
$$x = -7 + \frac{3 \cdot 5}{3}$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = 3$$
Rapid solution
$$x_{1} = -2$$
=
$$-2$$
=
$$y_{1} = 3$$
=
$$3$$
=
=
$$-2$$
=
-2
$$y_{1} = 3$$
=
$$3$$
=
3
Gaussian elimination
Given the system of equations
$$3 x - 5 y = -21$$
$$7 x + 10 y = 16$$
We give the system of equations to the canonical form
$$3 x - 5 y = -21$$
$$7 x + 10 y = 16$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -5 & -21\\7 & 10 & 16\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\7\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -5 & -21\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{3 \cdot 7}{3} & 10 - - \frac{35}{3} & 16 - - 49\end{matrix}\right] = \left[\begin{matrix}0 & \frac{65}{3} & 65\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -5 & -21\\0 & \frac{65}{3} & 65\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{65}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{65}{3} & 65\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-3\right) 0}{13} & -5 - \frac{\left(-3\right) 65}{3 \cdot 13} & -21 - \frac{\left(-3\right) 65}{13}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & -6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -6\\0 & \frac{65}{3} & 65\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} + 6 = 0$$
$$\frac{65 x_{2}}{3} - 65 = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = 3$$
$$3 x - 5 y = -21$$
$$7 x + 10 y = 16$$
We give the system of equations to the canonical form
$$3 x - 5 y = -21$$
$$7 x + 10 y = 16$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -5 & -21\\7 & 10 & 16\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\7\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -5 & -21\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{3 \cdot 7}{3} & 10 - - \frac{35}{3} & 16 - - 49\end{matrix}\right] = \left[\begin{matrix}0 & \frac{65}{3} & 65\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -5 & -21\\0 & \frac{65}{3} & 65\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{65}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{65}{3} & 65\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-3\right) 0}{13} & -5 - \frac{\left(-3\right) 65}{3 \cdot 13} & -21 - \frac{\left(-3\right) 65}{13}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & -6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -6\\0 & \frac{65}{3} & 65\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} + 6 = 0$$
$$\frac{65 x_{2}}{3} - 65 = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = 3$$
Cramer's rule
$$3 x - 5 y = -21$$
$$7 x + 10 y = 16$$
We give the system of equations to the canonical form
$$3 x - 5 y = -21$$
$$7 x + 10 y = 16$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 5 x_{2}\\7 x_{1} + 10 x_{2}\end{matrix}\right] = \left[\begin{matrix}-21\\16\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -5\\7 & 10\end{matrix}\right] \right)} = 65$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-21 & -5\\16 & 10\end{matrix}\right] \right)}}{65} = -2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -21\\7 & 16\end{matrix}\right] \right)}}{65} = 3$$
$$7 x + 10 y = 16$$
We give the system of equations to the canonical form
$$3 x - 5 y = -21$$
$$7 x + 10 y = 16$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 5 x_{2}\\7 x_{1} + 10 x_{2}\end{matrix}\right] = \left[\begin{matrix}-21\\16\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -5\\7 & 10\end{matrix}\right] \right)} = 65$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-21 & -5\\16 & 10\end{matrix}\right] \right)}}{65} = -2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -21\\7 & 16\end{matrix}\right] \right)}}{65} = 3$$