Mister Exam
4x+5y=11; 6x+8y=15
The solution
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4*x + 5*y = 11
$$4 x + 5 y = 11$$
6*x + 8*y = 15
$$6 x + 8 y = 15$$
6*x + 8*y = 15
Detail solution
Given the system of equations
$$4 x + 5 y = 11$$
$$6 x + 8 y = 15$$
Let's express from equation 1 x
$$4 x + 5 y = 11$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 11 - 5 y$$
$$4 x = 11 - 5 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{11 - 5 y}{4}$$
$$x = \frac{11}{4} - \frac{5 y}{4}$$
Let's try the obtained element x to 2-th equation
$$6 x + 8 y = 15$$
We get:
$$8 y + 6 \left(\frac{11}{4} - \frac{5 y}{4}\right) = 15$$
$$\frac{y}{2} + \frac{33}{2} = 15$$
We move the free summand 33/2 from the left part to the right part performing the sign change
$$\frac{y}{2} = - \frac{33}{2} + 15$$
$$\frac{y}{2} = - \frac{3}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$y = -3$$
Because
$$x = \frac{11}{4} - \frac{5 y}{4}$$
then
$$x = \frac{11}{4} - - \frac{15}{4}$$
$$x = \frac{13}{2}$$
The answer:
$$x = \frac{13}{2}$$
$$y = -3$$
$$4 x + 5 y = 11$$
$$6 x + 8 y = 15$$
Let's express from equation 1 x
$$4 x + 5 y = 11$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 11 - 5 y$$
$$4 x = 11 - 5 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{11 - 5 y}{4}$$
$$x = \frac{11}{4} - \frac{5 y}{4}$$
Let's try the obtained element x to 2-th equation
$$6 x + 8 y = 15$$
We get:
$$8 y + 6 \left(\frac{11}{4} - \frac{5 y}{4}\right) = 15$$
$$\frac{y}{2} + \frac{33}{2} = 15$$
We move the free summand 33/2 from the left part to the right part performing the sign change
$$\frac{y}{2} = - \frac{33}{2} + 15$$
$$\frac{y}{2} = - \frac{3}{2}$$
Let's divide both parts of the equation by the multiplier of y
/y\ |-| \2/ -3 --- = ----- 1/2 2*1/2
$$y = -3$$
Because
$$x = \frac{11}{4} - \frac{5 y}{4}$$
then
$$x = \frac{11}{4} - - \frac{15}{4}$$
$$x = \frac{13}{2}$$
The answer:
$$x = \frac{13}{2}$$
$$y = -3$$
Rapid solution
$$x_{1} = \frac{13}{2}$$
=
$$\frac{13}{2}$$
=
$$y_{1} = -3$$
=
$$-3$$
=
=
$$\frac{13}{2}$$
=
6.5
$$y_{1} = -3$$
=
$$-3$$
=
-3
Gaussian elimination
Given the system of equations
$$4 x + 5 y = 11$$
$$6 x + 8 y = 15$$
We give the system of equations to the canonical form
$$4 x + 5 y = 11$$
$$6 x + 8 y = 15$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 5 & 11\\6 & 8 & 15\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 5 & 11\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{3 \cdot 4}{2} & 8 - \frac{3 \cdot 5}{2} & 15 - \frac{3 \cdot 11}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{2} & - \frac{3}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 5 & 11\\0 & \frac{1}{2} & - \frac{3}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}5\\\frac{1}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{2} & - \frac{3}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - 0 \cdot 10 & 5 - \frac{10}{2} & 11 - \frac{\left(-3\right) 10}{2}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 26\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 26\\0 & \frac{1}{2} & - \frac{3}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 26 = 0$$
$$\frac{x_{2}}{2} + \frac{3}{2} = 0$$
We get the answer:
$$x_{1} = \frac{13}{2}$$
$$x_{2} = -3$$
$$4 x + 5 y = 11$$
$$6 x + 8 y = 15$$
We give the system of equations to the canonical form
$$4 x + 5 y = 11$$
$$6 x + 8 y = 15$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 5 & 11\\6 & 8 & 15\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 5 & 11\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{3 \cdot 4}{2} & 8 - \frac{3 \cdot 5}{2} & 15 - \frac{3 \cdot 11}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{2} & - \frac{3}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 5 & 11\\0 & \frac{1}{2} & - \frac{3}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}5\\\frac{1}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{2} & - \frac{3}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - 0 \cdot 10 & 5 - \frac{10}{2} & 11 - \frac{\left(-3\right) 10}{2}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 26\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 26\\0 & \frac{1}{2} & - \frac{3}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 26 = 0$$
$$\frac{x_{2}}{2} + \frac{3}{2} = 0$$
We get the answer:
$$x_{1} = \frac{13}{2}$$
$$x_{2} = -3$$
Cramer's rule
$$4 x + 5 y = 11$$
$$6 x + 8 y = 15$$
We give the system of equations to the canonical form
$$4 x + 5 y = 11$$
$$6 x + 8 y = 15$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + 5 x_{2}\\6 x_{1} + 8 x_{2}\end{matrix}\right] = \left[\begin{matrix}11\\15\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 5\\6 & 8\end{matrix}\right] \right)} = 2$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}11 & 5\\15 & 8\end{matrix}\right] \right)}}{2} = \frac{13}{2}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 11\\6 & 15\end{matrix}\right] \right)}}{2} = -3$$
$$6 x + 8 y = 15$$
We give the system of equations to the canonical form
$$4 x + 5 y = 11$$
$$6 x + 8 y = 15$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + 5 x_{2}\\6 x_{1} + 8 x_{2}\end{matrix}\right] = \left[\begin{matrix}11\\15\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 5\\6 & 8\end{matrix}\right] \right)} = 2$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}11 & 5\\15 & 8\end{matrix}\right] \right)}}{2} = \frac{13}{2}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 11\\6 & 15\end{matrix}\right] \right)}}{2} = -3$$