Mister Exam
3x+10y=12; 12x-5y=3
The solution
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3*x + 10*y = 12
$$3 x + 10 y = 12$$
12*x - 5*y = 3
$$12 x - 5 y = 3$$
12*x - 5*y = 3
Detail solution
Given the system of equations
$$3 x + 10 y = 12$$
$$12 x - 5 y = 3$$
Let's express from equation 1 x
$$3 x + 10 y = 12$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 12 - 10 y$$
$$3 x = 12 - 10 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{12 - 10 y}{3}$$
$$x = 4 - \frac{10 y}{3}$$
Let's try the obtained element x to 2-th equation
$$12 x - 5 y = 3$$
We get:
$$- 5 y + 12 \left(4 - \frac{10 y}{3}\right) = 3$$
$$48 - 45 y = 3$$
We move the free summand 48 from the left part to the right part performing the sign change
$$- 45 y = -48 + 3$$
$$- 45 y = -45$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 45 y}{-45} = - \frac{45}{-45}$$
$$y = 1$$
Because
$$x = 4 - \frac{10 y}{3}$$
then
$$x = - \frac{10}{3} + 4$$
$$x = \frac{2}{3}$$
The answer:
$$x = \frac{2}{3}$$
$$y = 1$$
$$3 x + 10 y = 12$$
$$12 x - 5 y = 3$$
Let's express from equation 1 x
$$3 x + 10 y = 12$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 12 - 10 y$$
$$3 x = 12 - 10 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{12 - 10 y}{3}$$
$$x = 4 - \frac{10 y}{3}$$
Let's try the obtained element x to 2-th equation
$$12 x - 5 y = 3$$
We get:
$$- 5 y + 12 \left(4 - \frac{10 y}{3}\right) = 3$$
$$48 - 45 y = 3$$
We move the free summand 48 from the left part to the right part performing the sign change
$$- 45 y = -48 + 3$$
$$- 45 y = -45$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 45 y}{-45} = - \frac{45}{-45}$$
$$y = 1$$
Because
$$x = 4 - \frac{10 y}{3}$$
then
$$x = - \frac{10}{3} + 4$$
$$x = \frac{2}{3}$$
The answer:
$$x = \frac{2}{3}$$
$$y = 1$$
Rapid solution
$$x_{1} = \frac{2}{3}$$
=
$$\frac{2}{3}$$
=
$$y_{1} = 1$$
=
$$1$$
=
=
$$\frac{2}{3}$$
=
0.666666666666667
$$y_{1} = 1$$
=
$$1$$
=
1
Cramer's rule
$$3 x + 10 y = 12$$
$$12 x - 5 y = 3$$
We give the system of equations to the canonical form
$$3 x + 10 y = 12$$
$$12 x - 5 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + 10 x_{2}\\12 x_{1} - 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}12\\3\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 10\\12 & -5\end{matrix}\right] \right)} = -135$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}12 & 10\\3 & -5\end{matrix}\right] \right)}}{135} = \frac{2}{3}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 12\\12 & 3\end{matrix}\right] \right)}}{135} = 1$$
$$12 x - 5 y = 3$$
We give the system of equations to the canonical form
$$3 x + 10 y = 12$$
$$12 x - 5 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + 10 x_{2}\\12 x_{1} - 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}12\\3\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 10\\12 & -5\end{matrix}\right] \right)} = -135$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}12 & 10\\3 & -5\end{matrix}\right] \right)}}{135} = \frac{2}{3}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 12\\12 & 3\end{matrix}\right] \right)}}{135} = 1$$
Gaussian elimination
Given the system of equations
$$3 x + 10 y = 12$$
$$12 x - 5 y = 3$$
We give the system of equations to the canonical form
$$3 x + 10 y = 12$$
$$12 x - 5 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 10 & 12\\12 & -5 & 3\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\12\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 10 & 12\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}12 - 3 \cdot 4 & - 4 \cdot 10 - 5 & 3 - 4 \cdot 12\end{matrix}\right] = \left[\begin{matrix}0 & -45 & -45\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 10 & 12\\0 & -45 & -45\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}10\\-45\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -45 & -45\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-2\right) 0}{9} & 10 - - -10 & 12 - - -10\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 2\\0 & -45 & -45\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 2 = 0$$
$$45 - 45 x_{2} = 0$$
We get the answer:
$$x_{1} = \frac{2}{3}$$
$$x_{2} = 1$$
$$3 x + 10 y = 12$$
$$12 x - 5 y = 3$$
We give the system of equations to the canonical form
$$3 x + 10 y = 12$$
$$12 x - 5 y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 10 & 12\\12 & -5 & 3\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\12\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 10 & 12\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}12 - 3 \cdot 4 & - 4 \cdot 10 - 5 & 3 - 4 \cdot 12\end{matrix}\right] = \left[\begin{matrix}0 & -45 & -45\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 10 & 12\\0 & -45 & -45\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}10\\-45\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -45 & -45\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-2\right) 0}{9} & 10 - - -10 & 12 - - -10\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 2\\0 & -45 & -45\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 2 = 0$$
$$45 - 45 x_{2} = 0$$
We get the answer:
$$x_{1} = \frac{2}{3}$$
$$x_{2} = 1$$