Mister Exam
3x+2y=8; 4x−y=7
The solution
Detail solution
Given the system of equations
$$3 x + 2 y = 8$$
$$4 x - y = 7$$
Let's express from equation 1 x
$$3 x + 2 y = 8$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 8 - 2 y$$
$$3 x = 8 - 2 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{8 - 2 y}{3}$$
$$x = \frac{8}{3} - \frac{2 y}{3}$$
Let's try the obtained element x to 2-th equation
$$4 x - y = 7$$
We get:
$$- y + 4 \left(\frac{8}{3} - \frac{2 y}{3}\right) = 7$$
$$\frac{32}{3} - \frac{11 y}{3} = 7$$
We move the free summand 32/3 from the left part to the right part performing the sign change
$$- \frac{11 y}{3} = - \frac{32}{3} + 7$$
$$- \frac{11 y}{3} = - \frac{11}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{11}{3} y}{- \frac{11}{3}} = - \frac{11}{\left(- \frac{11}{3}\right) 3}$$
$$y = 1$$
Because
$$x = \frac{8}{3} - \frac{2 y}{3}$$
then
$$x = - \frac{2}{3} + \frac{8}{3}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = 1$$
$$3 x + 2 y = 8$$
$$4 x - y = 7$$
Let's express from equation 1 x
$$3 x + 2 y = 8$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 8 - 2 y$$
$$3 x = 8 - 2 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{8 - 2 y}{3}$$
$$x = \frac{8}{3} - \frac{2 y}{3}$$
Let's try the obtained element x to 2-th equation
$$4 x - y = 7$$
We get:
$$- y + 4 \left(\frac{8}{3} - \frac{2 y}{3}\right) = 7$$
$$\frac{32}{3} - \frac{11 y}{3} = 7$$
We move the free summand 32/3 from the left part to the right part performing the sign change
$$- \frac{11 y}{3} = - \frac{32}{3} + 7$$
$$- \frac{11 y}{3} = - \frac{11}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{11}{3} y}{- \frac{11}{3}} = - \frac{11}{\left(- \frac{11}{3}\right) 3}$$
$$y = 1$$
Because
$$x = \frac{8}{3} - \frac{2 y}{3}$$
then
$$x = - \frac{2}{3} + \frac{8}{3}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = 1$$
Rapid solution
$$x_{1} = 2$$
=
$$2$$
=
$$y_{1} = 1$$
=
$$1$$
=
=
$$2$$
=
2
$$y_{1} = 1$$
=
$$1$$
=
1
Gaussian elimination
Given the system of equations
$$3 x + 2 y = 8$$
$$4 x - y = 7$$
We give the system of equations to the canonical form
$$3 x + 2 y = 8$$
$$4 x - y = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 2 & 8\\4 & -1 & 7\end{matrix}\right]$$
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
$$\left[\begin{matrix}3 & 2 & 8\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{3 \cdot 4}{3} & - \frac{2 \cdot 4}{3} - 1 & 7 - \frac{4 \cdot 8}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{11}{3} & - \frac{11}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 2 & 8\\0 & - \frac{11}{3} & - \frac{11}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}2\\- \frac{11}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{11}{3} & - \frac{11}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-6\right) 0}{11} & 2 - - -2 & 8 - - -2\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 6\\0 & - \frac{11}{3} & - \frac{11}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 6 = 0$$
$$\frac{11}{3} - \frac{11 x_{2}}{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = 1$$
$$3 x + 2 y = 8$$
$$4 x - y = 7$$
We give the system of equations to the canonical form
$$3 x + 2 y = 8$$
$$4 x - y = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 2 & 8\\4 & -1 & 7\end{matrix}\right]$$
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
$$\left[\begin{matrix}3 & 2 & 8\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{3 \cdot 4}{3} & - \frac{2 \cdot 4}{3} - 1 & 7 - \frac{4 \cdot 8}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{11}{3} & - \frac{11}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 2 & 8\\0 & - \frac{11}{3} & - \frac{11}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}2\\- \frac{11}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{11}{3} & - \frac{11}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-6\right) 0}{11} & 2 - - -2 & 8 - - -2\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 6\\0 & - \frac{11}{3} & - \frac{11}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 6 = 0$$
$$\frac{11}{3} - \frac{11 x_{2}}{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = 1$$
Cramer's rule
$$3 x + 2 y = 8$$
$$4 x - y = 7$$
We give the system of equations to the canonical form
$$3 x + 2 y = 8$$
$$4 x - y = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + 2 x_{2}\\4 x_{1} - x_{2}\end{matrix}\right] = \left[\begin{matrix}8\\7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 2\\4 & -1\end{matrix}\right] \right)} = -11$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}8 & 2\\7 & -1\end{matrix}\right] \right)}}{11} = 2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 8\\4 & 7\end{matrix}\right] \right)}}{11} = 1$$
$$4 x - y = 7$$
We give the system of equations to the canonical form
$$3 x + 2 y = 8$$
$$4 x - y = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + 2 x_{2}\\4 x_{1} - x_{2}\end{matrix}\right] = \left[\begin{matrix}8\\7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 2\\4 & -1\end{matrix}\right] \right)} = -11$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}8 & 2\\7 & -1\end{matrix}\right] \right)}}{11} = 2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 8\\4 & 7\end{matrix}\right] \right)}}{11} = 1$$