Mister Exam
2x-3y=2; 4x-6y=3
The solution
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2*x - 3*y = 2
$$2 x - 3 y = 2$$
4*x - 6*y = 3
$$4 x - 6 y = 3$$
4*x - 6*y = 3
Detail solution
Given the system of equations
$$2 x - 3 y = 2$$
$$4 x - 6 y = 3$$
Let's express from equation 1 x
$$2 x - 3 y = 2$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 3 y + 2$$
$$2 x = 3 y + 2$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{3 y + 2}{2}$$
$$x = \frac{3 y}{2} + 1$$
Let's try the obtained element x to 2-th equation
$$4 x - 6 y = 3$$
We get:
$$- 6 y + 4 \left(\frac{3 y}{2} + 1\right) = 3$$
so
This system of equations has no solutions
$$2 x - 3 y = 2$$
$$4 x - 6 y = 3$$
Let's express from equation 1 x
$$2 x - 3 y = 2$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 3 y + 2$$
$$2 x = 3 y + 2$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{3 y + 2}{2}$$
$$x = \frac{3 y}{2} + 1$$
Let's try the obtained element x to 2-th equation
$$4 x - 6 y = 3$$
We get:
$$- 6 y + 4 \left(\frac{3 y}{2} + 1\right) = 3$$
so
This system of equations has no solutions
Gaussian elimination
Given the system of equations
$$2 x - 3 y = 2$$
$$4 x - 6 y = 3$$
We give the system of equations to the canonical form
$$2 x - 3 y = 2$$
$$4 x - 6 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -3 & 2\\4 & -6 & 3\end{matrix}\right]$$
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
$$\left[\begin{matrix}2 & -3 & 2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - 2 \cdot 2 & -6 - - 6 & 3 - 2 \cdot 2\end{matrix}\right] = \left[\begin{matrix}0 & 0 & -1\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -3 & 2\\0 & 0 & -1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -3 & 2\end{matrix}\right]$$
,
and subtract it from other lines:
We prepare elementary equations using a solved matrix and see that this system of equations has no decisions
$$2 x_{1} - 3 x_{2} - 2 = 0$$
$$0 + 1 = 0$$
We get the answer:
This system of equations has no solutions
$$2 x - 3 y = 2$$
$$4 x - 6 y = 3$$
We give the system of equations to the canonical form
$$2 x - 3 y = 2$$
$$4 x - 6 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -3 & 2\\4 & -6 & 3\end{matrix}\right]$$
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
$$\left[\begin{matrix}2 & -3 & 2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - 2 \cdot 2 & -6 - - 6 & 3 - 2 \cdot 2\end{matrix}\right] = \left[\begin{matrix}0 & 0 & -1\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -3 & 2\\0 & 0 & -1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -3 & 2\end{matrix}\right]$$
,
and subtract it from other lines:
We prepare elementary equations using a solved matrix and see that this system of equations has no decisions
$$2 x_{1} - 3 x_{2} - 2 = 0$$
$$0 + 1 = 0$$
We get the answer:
This system of equations has no solutions