Mister Exam
3x-7y=32; X=-5y-4
The solution
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3*x - 7*y = 32
$$3 x - 7 y = 32$$
x = -5*y - 4
$$x = - 5 y - 4$$
x = -5*y - 4
Detail solution
Given the system of equations
$$3 x - 7 y = 32$$
$$x = - 5 y - 4$$
Let's express from equation 1 x
$$3 x - 7 y = 32$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 7 y + 32$$
$$3 x = 7 y + 32$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{7 y + 32}{3}$$
$$x = \frac{7 y}{3} + \frac{32}{3}$$
Let's try the obtained element x to 2-th equation
$$x = - 5 y - 4$$
We get:
$$\frac{7 y}{3} + \frac{32}{3} = - 5 y - 4$$
$$\frac{7 y}{3} + \frac{32}{3} = - 5 y - 4$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$5 y + \left(\frac{7 y}{3} + \frac{32}{3}\right) = -4$$
$$\frac{22 y}{3} + \frac{32}{3} = -4$$
We move the free summand 32/3 from the left part to the right part performing the sign change
$$\frac{22 y}{3} = - \frac{32}{3} - 4$$
$$\frac{22 y}{3} = - \frac{44}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{22}{3} y}{\frac{22}{3}} = - \frac{44}{3 \frac{22}{3}}$$
$$y = -2$$
Because
$$x = \frac{7 y}{3} + \frac{32}{3}$$
then
$$x = \frac{\left(-2\right) 7}{3} + \frac{32}{3}$$
$$x = 6$$
The answer:
$$x = 6$$
$$y = -2$$
$$3 x - 7 y = 32$$
$$x = - 5 y - 4$$
Let's express from equation 1 x
$$3 x - 7 y = 32$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 7 y + 32$$
$$3 x = 7 y + 32$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{7 y + 32}{3}$$
$$x = \frac{7 y}{3} + \frac{32}{3}$$
Let's try the obtained element x to 2-th equation
$$x = - 5 y - 4$$
We get:
$$\frac{7 y}{3} + \frac{32}{3} = - 5 y - 4$$
$$\frac{7 y}{3} + \frac{32}{3} = - 5 y - 4$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$5 y + \left(\frac{7 y}{3} + \frac{32}{3}\right) = -4$$
$$\frac{22 y}{3} + \frac{32}{3} = -4$$
We move the free summand 32/3 from the left part to the right part performing the sign change
$$\frac{22 y}{3} = - \frac{32}{3} - 4$$
$$\frac{22 y}{3} = - \frac{44}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{22}{3} y}{\frac{22}{3}} = - \frac{44}{3 \frac{22}{3}}$$
$$y = -2$$
Because
$$x = \frac{7 y}{3} + \frac{32}{3}$$
then
$$x = \frac{\left(-2\right) 7}{3} + \frac{32}{3}$$
$$x = 6$$
The answer:
$$x = 6$$
$$y = -2$$
Rapid solution
$$x_{1} = 6$$
=
$$6$$
=
$$y_{1} = -2$$
=
$$-2$$
=
=
$$6$$
=
6
$$y_{1} = -2$$
=
$$-2$$
=
-2
Gaussian elimination
Given the system of equations
$$3 x - 7 y = 32$$
$$x = - 5 y - 4$$
We give the system of equations to the canonical form
$$3 x - 7 y = 32$$
$$x + 5 y = -4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -7 & 32\\1 & 5 & -4\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -7 & 32\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{3}{3} & 5 - - \frac{7}{3} & - \frac{32}{3} - 4\end{matrix}\right] = \left[\begin{matrix}0 & \frac{22}{3} & - \frac{44}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -7 & 32\\0 & \frac{22}{3} & - \frac{44}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-7\\\frac{22}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{22}{3} & - \frac{44}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-21\right) 0}{22} & -7 - \frac{\left(-21\right) 22}{3 \cdot 22} & 32 - - -14\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 18\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 18\\0 & \frac{22}{3} & - \frac{44}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 18 = 0$$
$$\frac{22 x_{2}}{3} + \frac{44}{3} = 0$$
We get the answer:
$$x_{1} = 6$$
$$x_{2} = -2$$
$$3 x - 7 y = 32$$
$$x = - 5 y - 4$$
We give the system of equations to the canonical form
$$3 x - 7 y = 32$$
$$x + 5 y = -4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -7 & 32\\1 & 5 & -4\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -7 & 32\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{3}{3} & 5 - - \frac{7}{3} & - \frac{32}{3} - 4\end{matrix}\right] = \left[\begin{matrix}0 & \frac{22}{3} & - \frac{44}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -7 & 32\\0 & \frac{22}{3} & - \frac{44}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-7\\\frac{22}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{22}{3} & - \frac{44}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-21\right) 0}{22} & -7 - \frac{\left(-21\right) 22}{3 \cdot 22} & 32 - - -14\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 18\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 18\\0 & \frac{22}{3} & - \frac{44}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 18 = 0$$
$$\frac{22 x_{2}}{3} + \frac{44}{3} = 0$$
We get the answer:
$$x_{1} = 6$$
$$x_{2} = -2$$
Cramer's rule
$$3 x - 7 y = 32$$
$$x = - 5 y - 4$$
We give the system of equations to the canonical form
$$3 x - 7 y = 32$$
$$x + 5 y = -4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 7 x_{2}\\x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}32\\-4\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -7\\1 & 5\end{matrix}\right] \right)} = 22$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}32 & -7\\-4 & 5\end{matrix}\right] \right)}}{22} = 6$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 32\\1 & -4\end{matrix}\right] \right)}}{22} = -2$$
$$x = - 5 y - 4$$
We give the system of equations to the canonical form
$$3 x - 7 y = 32$$
$$x + 5 y = -4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 7 x_{2}\\x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}32\\-4\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -7\\1 & 5\end{matrix}\right] \right)} = 22$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}32 & -7\\-4 & 5\end{matrix}\right] \right)}}{22} = 6$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 32\\1 & -4\end{matrix}\right] \right)}}{22} = -2$$