Mister Exam
2x+3y=5; 4x-4y=-1
The solution
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2*x + 3*y = 5
$$2 x + 3 y = 5$$
4*x - 4*y = -1
$$4 x - 4 y = -1$$
4*x - 4*y = -1
Detail solution
Given the system of equations
$$2 x + 3 y = 5$$
$$4 x - 4 y = -1$$
Let's express from equation 1 x
$$2 x + 3 y = 5$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 5 - 3 y$$
$$2 x = 5 - 3 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{5 - 3 y}{2}$$
$$x = \frac{5}{2} - \frac{3 y}{2}$$
Let's try the obtained element x to 2-th equation
$$4 x - 4 y = -1$$
We get:
$$- 4 y + 4 \left(\frac{5}{2} - \frac{3 y}{2}\right) = -1$$
$$10 - 10 y = -1$$
We move the free summand 10 from the left part to the right part performing the sign change
$$- 10 y = -10 - 1$$
$$- 10 y = -11$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 10 y}{-10} = - \frac{11}{-10}$$
$$y = \frac{11}{10}$$
Because
$$x = \frac{5}{2} - \frac{3 y}{2}$$
then
$$x = \frac{5}{2} - \frac{33}{20}$$
$$x = \frac{17}{20}$$
The answer:
$$x = \frac{17}{20}$$
$$y = \frac{11}{10}$$
$$2 x + 3 y = 5$$
$$4 x - 4 y = -1$$
Let's express from equation 1 x
$$2 x + 3 y = 5$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 5 - 3 y$$
$$2 x = 5 - 3 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{5 - 3 y}{2}$$
$$x = \frac{5}{2} - \frac{3 y}{2}$$
Let's try the obtained element x to 2-th equation
$$4 x - 4 y = -1$$
We get:
$$- 4 y + 4 \left(\frac{5}{2} - \frac{3 y}{2}\right) = -1$$
$$10 - 10 y = -1$$
We move the free summand 10 from the left part to the right part performing the sign change
$$- 10 y = -10 - 1$$
$$- 10 y = -11$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 10 y}{-10} = - \frac{11}{-10}$$
$$y = \frac{11}{10}$$
Because
$$x = \frac{5}{2} - \frac{3 y}{2}$$
then
$$x = \frac{5}{2} - \frac{33}{20}$$
$$x = \frac{17}{20}$$
The answer:
$$x = \frac{17}{20}$$
$$y = \frac{11}{10}$$
Rapid solution
$$x_{1} = \frac{17}{20}$$
=
$$\frac{17}{20}$$
=
$$y_{1} = \frac{11}{10}$$
=
$$\frac{11}{10}$$
=
=
$$\frac{17}{20}$$
=
0.85
$$y_{1} = \frac{11}{10}$$
=
$$\frac{11}{10}$$
=
1.1
Gaussian elimination
Given the system of equations
$$2 x + 3 y = 5$$
$$4 x - 4 y = -1$$
We give the system of equations to the canonical form
$$2 x + 3 y = 5$$
$$4 x - 4 y = -1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 3 & 5\\4 & -4 & -1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 3 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - 2 \cdot 2 & - 2 \cdot 3 - 4 & - 2 \cdot 5 - 1\end{matrix}\right] = \left[\begin{matrix}0 & -10 & -11\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 3 & 5\\0 & -10 & -11\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}3\\-10\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -10 & -11\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-3\right) 0}{10} & 3 - - -3 & 5 - - \frac{-33}{10}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & \frac{17}{10}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & \frac{17}{10}\\0 & -10 & -11\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - \frac{17}{10} = 0$$
$$11 - 10 x_{2} = 0$$
We get the answer:
$$x_{1} = \frac{17}{20}$$
$$x_{2} = \frac{11}{10}$$
$$2 x + 3 y = 5$$
$$4 x - 4 y = -1$$
We give the system of equations to the canonical form
$$2 x + 3 y = 5$$
$$4 x - 4 y = -1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 3 & 5\\4 & -4 & -1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 3 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - 2 \cdot 2 & - 2 \cdot 3 - 4 & - 2 \cdot 5 - 1\end{matrix}\right] = \left[\begin{matrix}0 & -10 & -11\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 3 & 5\\0 & -10 & -11\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}3\\-10\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -10 & -11\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-3\right) 0}{10} & 3 - - -3 & 5 - - \frac{-33}{10}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & \frac{17}{10}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & \frac{17}{10}\\0 & -10 & -11\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - \frac{17}{10} = 0$$
$$11 - 10 x_{2} = 0$$
We get the answer:
$$x_{1} = \frac{17}{20}$$
$$x_{2} = \frac{11}{10}$$
Cramer's rule
$$2 x + 3 y = 5$$
$$4 x - 4 y = -1$$
We give the system of equations to the canonical form
$$2 x + 3 y = 5$$
$$4 x - 4 y = -1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + 3 x_{2}\\4 x_{1} - 4 x_{2}\end{matrix}\right] = \left[\begin{matrix}5\\-1\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 3\\4 & -4\end{matrix}\right] \right)} = -20$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 3\\-1 & -4\end{matrix}\right] \right)}}{20} = \frac{17}{20}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 5\\4 & -1\end{matrix}\right] \right)}}{20} = \frac{11}{10}$$
$$4 x - 4 y = -1$$
We give the system of equations to the canonical form
$$2 x + 3 y = 5$$
$$4 x - 4 y = -1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + 3 x_{2}\\4 x_{1} - 4 x_{2}\end{matrix}\right] = \left[\begin{matrix}5\\-1\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 3\\4 & -4\end{matrix}\right] \right)} = -20$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 3\\-1 & -4\end{matrix}\right] \right)}}{20} = \frac{17}{20}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 5\\4 & -1\end{matrix}\right] \right)}}{20} = \frac{11}{10}$$