Mister Exam
3x-2y=7; -4x+3y=-7
The solution
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3*x - 2*y = 7
$$3 x - 2 y = 7$$
-4*x + 3*y = -7
$$- 4 x + 3 y = -7$$
-4*x + 3*y = -7
Detail solution
Given the system of equations
$$3 x - 2 y = 7$$
$$- 4 x + 3 y = -7$$
Let's express from equation 1 x
$$3 x - 2 y = 7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 2 y + 7$$
$$3 x = 2 y + 7$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{2 y + 7}{3}$$
$$x = \frac{2 y}{3} + \frac{7}{3}$$
Let's try the obtained element x to 2-th equation
$$- 4 x + 3 y = -7$$
We get:
$$3 y - 4 \left(\frac{2 y}{3} + \frac{7}{3}\right) = -7$$
$$\frac{y}{3} - \frac{28}{3} = -7$$
We move the free summand -28/3 from the left part to the right part performing the sign change
$$\frac{y}{3} = -7 + \frac{28}{3}$$
$$\frac{y}{3} = \frac{7}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$y = 7$$
Because
$$x = \frac{2 y}{3} + \frac{7}{3}$$
then
$$x = \frac{7}{3} + \frac{2 \cdot 7}{3}$$
$$x = 7$$
The answer:
$$x = 7$$
$$y = 7$$
$$3 x - 2 y = 7$$
$$- 4 x + 3 y = -7$$
Let's express from equation 1 x
$$3 x - 2 y = 7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 2 y + 7$$
$$3 x = 2 y + 7$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{2 y + 7}{3}$$
$$x = \frac{2 y}{3} + \frac{7}{3}$$
Let's try the obtained element x to 2-th equation
$$- 4 x + 3 y = -7$$
We get:
$$3 y - 4 \left(\frac{2 y}{3} + \frac{7}{3}\right) = -7$$
$$\frac{y}{3} - \frac{28}{3} = -7$$
We move the free summand -28/3 from the left part to the right part performing the sign change
$$\frac{y}{3} = -7 + \frac{28}{3}$$
$$\frac{y}{3} = \frac{7}{3}$$
Let's divide both parts of the equation by the multiplier of y
/y\ |-| \3/ 7 --- = ----- 1/3 3*1/3
$$y = 7$$
Because
$$x = \frac{2 y}{3} + \frac{7}{3}$$
then
$$x = \frac{7}{3} + \frac{2 \cdot 7}{3}$$
$$x = 7$$
The answer:
$$x = 7$$
$$y = 7$$
Rapid solution
$$x_{1} = 7$$
=
$$7$$
=
$$y_{1} = 7$$
=
$$7$$
=
=
$$7$$
=
7
$$y_{1} = 7$$
=
$$7$$
=
7
Cramer's rule
$$3 x - 2 y = 7$$
$$- 4 x + 3 y = -7$$
We give the system of equations to the canonical form
$$3 x - 2 y = 7$$
$$- 4 x + 3 y = -7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 2 x_{2}\\- 4 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}7\\-7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -2\\-4 & 3\end{matrix}\right] \right)} = 1$$
, 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 )
$$x_{1} = \operatorname{det}{\left(\left[\begin{matrix}7 & -2\\-7 & 3\end{matrix}\right] \right)} = 7$$
$$x_{2} = \operatorname{det}{\left(\left[\begin{matrix}3 & 7\\-4 & -7\end{matrix}\right] \right)} = 7$$
$$- 4 x + 3 y = -7$$
We give the system of equations to the canonical form
$$3 x - 2 y = 7$$
$$- 4 x + 3 y = -7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 2 x_{2}\\- 4 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}7\\-7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -2\\-4 & 3\end{matrix}\right] \right)} = 1$$
, 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 )
$$x_{1} = \operatorname{det}{\left(\left[\begin{matrix}7 & -2\\-7 & 3\end{matrix}\right] \right)} = 7$$
$$x_{2} = \operatorname{det}{\left(\left[\begin{matrix}3 & 7\\-4 & -7\end{matrix}\right] \right)} = 7$$
Gaussian elimination
Given the system of equations
$$3 x - 2 y = 7$$
$$- 4 x + 3 y = -7$$
We give the system of equations to the canonical form
$$3 x - 2 y = 7$$
$$- 4 x + 3 y = -7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -2 & 7\\-4 & 3 & -7\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\-4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -2 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-4 - \frac{\left(-4\right) 3}{3} & 3 - - \frac{-8}{3} & -7 - \frac{\left(-4\right) 7}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{3} & \frac{7}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 7\\0 & \frac{1}{3} & \frac{7}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\\frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{3} & \frac{7}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \left(-6\right) 0 & -2 - - 2 & 7 - - 14\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 21\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 21\\0 & \frac{1}{3} & \frac{7}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 21 = 0$$
$$\frac{x_{2}}{3} - \frac{7}{3} = 0$$
We get the answer:
$$x_{1} = 7$$
$$x_{2} = 7$$
$$3 x - 2 y = 7$$
$$- 4 x + 3 y = -7$$
We give the system of equations to the canonical form
$$3 x - 2 y = 7$$
$$- 4 x + 3 y = -7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -2 & 7\\-4 & 3 & -7\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\-4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -2 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-4 - \frac{\left(-4\right) 3}{3} & 3 - - \frac{-8}{3} & -7 - \frac{\left(-4\right) 7}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{3} & \frac{7}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 7\\0 & \frac{1}{3} & \frac{7}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\\frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{3} & \frac{7}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \left(-6\right) 0 & -2 - - 2 & 7 - - 14\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 21\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 21\\0 & \frac{1}{3} & \frac{7}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 21 = 0$$
$$\frac{x_{2}}{3} - \frac{7}{3} = 0$$
We get the answer:
$$x_{1} = 7$$
$$x_{2} = 7$$