Mister Exam
1/4х-1/3у=4; 4/5х-3у=7
The solution
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x y - - - = 4 4 3
$$\frac{x}{4} - \frac{y}{3} = 4$$
4*x --- - 3*y = 7 5
$$\frac{4 x}{5} - 3 y = 7$$
4*x/5 - 3*y = 7
Detail solution
Given the system of equations
$$\frac{x}{4} - \frac{y}{3} = 4$$
$$\frac{4 x}{5} - 3 y = 7$$
Let's express from equation 1 x
$$\frac{x}{4} - \frac{y}{3} = 4$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{x}{4} = \frac{y}{3} + 4$$
$$\frac{x}{4} = \frac{y}{3} + 4$$
Let's divide both parts of the equation by the multiplier of x
$$x = \frac{4 y}{3} + 16$$
Let's try the obtained element x to 2-th equation
$$\frac{4 x}{5} - 3 y = 7$$
We get:
$$- 3 y + \frac{4 \left(\frac{4 y}{3} + 16\right)}{5} = 7$$
$$\frac{64}{5} - \frac{29 y}{15} = 7$$
We move the free summand 64/5 from the left part to the right part performing the sign change
$$- \frac{29 y}{15} = - \frac{64}{5} + 7$$
$$- \frac{29 y}{15} = - \frac{29}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{29}{15} y}{- \frac{29}{15}} = - \frac{29}{\left(- \frac{29}{15}\right) 5}$$
$$y = 3$$
Because
$$x = \frac{4 y}{3} + 16$$
then
$$x = \frac{3 \cdot 4}{3} + 16$$
$$x = 20$$
The answer:
$$x = 20$$
$$y = 3$$
$$\frac{x}{4} - \frac{y}{3} = 4$$
$$\frac{4 x}{5} - 3 y = 7$$
Let's express from equation 1 x
$$\frac{x}{4} - \frac{y}{3} = 4$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{x}{4} = \frac{y}{3} + 4$$
$$\frac{x}{4} = \frac{y}{3} + 4$$
Let's divide both parts of the equation by the multiplier of x
/x\ y |-| 4 + - \4/ 3 --- = ----- 1/4 1/4
$$x = \frac{4 y}{3} + 16$$
Let's try the obtained element x to 2-th equation
$$\frac{4 x}{5} - 3 y = 7$$
We get:
$$- 3 y + \frac{4 \left(\frac{4 y}{3} + 16\right)}{5} = 7$$
$$\frac{64}{5} - \frac{29 y}{15} = 7$$
We move the free summand 64/5 from the left part to the right part performing the sign change
$$- \frac{29 y}{15} = - \frac{64}{5} + 7$$
$$- \frac{29 y}{15} = - \frac{29}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{29}{15} y}{- \frac{29}{15}} = - \frac{29}{\left(- \frac{29}{15}\right) 5}$$
$$y = 3$$
Because
$$x = \frac{4 y}{3} + 16$$
then
$$x = \frac{3 \cdot 4}{3} + 16$$
$$x = 20$$
The answer:
$$x = 20$$
$$y = 3$$
Rapid solution
$$x_{1} = 20$$
=
$$20$$
=
$$y_{1} = 3$$
=
$$3$$
=
=
$$20$$
=
20
$$y_{1} = 3$$
=
$$3$$
=
3
Cramer's rule
$$\frac{x}{4} - \frac{y}{3} = 4$$
$$\frac{4 x}{5} - 3 y = 7$$
We give the system of equations to the canonical form
$$\frac{x}{4} - \frac{y}{3} = 4$$
$$\frac{4 x}{5} - 3 y = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{x_{1}}{4} - \frac{x_{2}}{3}\\\frac{4 x_{1}}{5} - 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}4\\7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{4} & - \frac{1}{3}\\\frac{4}{5} & -3\end{matrix}\right] \right)} = - \frac{29}{60}$$
, 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 )
$$x_{1} = - \frac{60 \operatorname{det}{\left(\left[\begin{matrix}4 & - \frac{1}{3}\\7 & -3\end{matrix}\right] \right)}}{29} = 20$$
$$x_{2} = - \frac{60 \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{4} & 4\\\frac{4}{5} & 7\end{matrix}\right] \right)}}{29} = 3$$
$$\frac{4 x}{5} - 3 y = 7$$
We give the system of equations to the canonical form
$$\frac{x}{4} - \frac{y}{3} = 4$$
$$\frac{4 x}{5} - 3 y = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{x_{1}}{4} - \frac{x_{2}}{3}\\\frac{4 x_{1}}{5} - 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}4\\7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{4} & - \frac{1}{3}\\\frac{4}{5} & -3\end{matrix}\right] \right)} = - \frac{29}{60}$$
, 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 )
$$x_{1} = - \frac{60 \operatorname{det}{\left(\left[\begin{matrix}4 & - \frac{1}{3}\\7 & -3\end{matrix}\right] \right)}}{29} = 20$$
$$x_{2} = - \frac{60 \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{4} & 4\\\frac{4}{5} & 7\end{matrix}\right] \right)}}{29} = 3$$
Gaussian elimination
Given the system of equations
$$\frac{x}{4} - \frac{y}{3} = 4$$
$$\frac{4 x}{5} - 3 y = 7$$
We give the system of equations to the canonical form
$$\frac{x}{4} - \frac{y}{3} = 4$$
$$\frac{4 x}{5} - 3 y = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{1}{4} & - \frac{1}{3} & 4\\\frac{4}{5} & -3 & 7\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{1}{4}\\\frac{4}{5}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{1}{4} & - \frac{1}{3} & 4\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{4}{5} - \frac{16}{4 \cdot 5} & -3 - - \frac{16}{15} & 7 - \frac{4 \cdot 16}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{29}{15} & - \frac{29}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{1}{4} & - \frac{1}{3} & 4\\0 & - \frac{29}{15} & - \frac{29}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{3}\\- \frac{29}{15}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{29}{15} & - \frac{29}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{1}{4} - \frac{0 \cdot 5}{29} & - \frac{1}{3} - - \frac{1}{3} & 4 - - 1\end{matrix}\right] = \left[\begin{matrix}\frac{1}{4} & 0 & 5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{1}{4} & 0 & 5\\0 & - \frac{29}{15} & - \frac{29}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{x_{1}}{4} - 5 = 0$$
$$\frac{29}{5} - \frac{29 x_{2}}{15} = 0$$
We get the answer:
$$x_{1} = 20$$
$$x_{2} = 3$$
$$\frac{x}{4} - \frac{y}{3} = 4$$
$$\frac{4 x}{5} - 3 y = 7$$
We give the system of equations to the canonical form
$$\frac{x}{4} - \frac{y}{3} = 4$$
$$\frac{4 x}{5} - 3 y = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{1}{4} & - \frac{1}{3} & 4\\\frac{4}{5} & -3 & 7\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{1}{4}\\\frac{4}{5}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{1}{4} & - \frac{1}{3} & 4\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{4}{5} - \frac{16}{4 \cdot 5} & -3 - - \frac{16}{15} & 7 - \frac{4 \cdot 16}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{29}{15} & - \frac{29}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{1}{4} & - \frac{1}{3} & 4\\0 & - \frac{29}{15} & - \frac{29}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{3}\\- \frac{29}{15}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{29}{15} & - \frac{29}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{1}{4} - \frac{0 \cdot 5}{29} & - \frac{1}{3} - - \frac{1}{3} & 4 - - 1\end{matrix}\right] = \left[\begin{matrix}\frac{1}{4} & 0 & 5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{1}{4} & 0 & 5\\0 & - \frac{29}{15} & - \frac{29}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{x_{1}}{4} - 5 = 0$$
$$\frac{29}{5} - \frac{29 x_{2}}{15} = 0$$
We get the answer:
$$x_{1} = 20$$
$$x_{2} = 3$$