Mister Exam
5p-3q=0; 3p+4q=29
The solution
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5*p - 3*q = 0
$$5 p - 3 q = 0$$
3*p + 4*q = 29
$$3 p + 4 q = 29$$
3*p + 4*q = 29
Detail solution
Given the system of equations
$$5 p - 3 q = 0$$
$$3 p + 4 q = 29$$
Let's express from equation 1 p
$$5 p - 3 q = 0$$
Let's move the summand with the variable q from the left part to the right part performing the sign change
$$5 p = 3 q$$
$$5 p = 3 q$$
Let's divide both parts of the equation by the multiplier of p
$$\frac{5 p}{5} = \frac{3 q}{5}$$
$$p = \frac{3 q}{5}$$
Let's try the obtained element p to 2-th equation
$$3 p + 4 q = 29$$
We get:
$$3 \frac{3 q}{5} + 4 q = 29$$
$$\frac{29 q}{5} = 29$$
Let's divide both parts of the equation by the multiplier of q
$$\frac{\frac{29}{5} q}{\frac{29}{5}} = \frac{29}{\frac{29}{5}}$$
$$q = 5$$
Because
$$p = \frac{3 q}{5}$$
then
$$p = \frac{3 \cdot 5}{5}$$
$$p = 3$$
The answer:
$$p = 3$$
$$q = 5$$
$$5 p - 3 q = 0$$
$$3 p + 4 q = 29$$
Let's express from equation 1 p
$$5 p - 3 q = 0$$
Let's move the summand with the variable q from the left part to the right part performing the sign change
$$5 p = 3 q$$
$$5 p = 3 q$$
Let's divide both parts of the equation by the multiplier of p
$$\frac{5 p}{5} = \frac{3 q}{5}$$
$$p = \frac{3 q}{5}$$
Let's try the obtained element p to 2-th equation
$$3 p + 4 q = 29$$
We get:
$$3 \frac{3 q}{5} + 4 q = 29$$
$$\frac{29 q}{5} = 29$$
Let's divide both parts of the equation by the multiplier of q
$$\frac{\frac{29}{5} q}{\frac{29}{5}} = \frac{29}{\frac{29}{5}}$$
$$q = 5$$
Because
$$p = \frac{3 q}{5}$$
then
$$p = \frac{3 \cdot 5}{5}$$
$$p = 3$$
The answer:
$$p = 3$$
$$q = 5$$
Rapid solution
$$p_{1} = 3$$
=
$$3$$
=
$$q_{1} = 5$$
=
$$5$$
=
=
$$3$$
=
3
$$q_{1} = 5$$
=
$$5$$
=
5
Cramer's rule
$$5 p - 3 q = 0$$
$$3 p + 4 q = 29$$
We give the system of equations to the canonical form
$$5 p - 3 q = 0$$
$$3 p + 4 q = 29$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 3 x_{2}\\3 x_{1} + 4 x_{2}\end{matrix}\right] = \left[\begin{matrix}0\\29\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -3\\3 & 4\end{matrix}\right] \right)} = 29$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}0 & -3\\29 & 4\end{matrix}\right] \right)}}{29} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 0\\3 & 29\end{matrix}\right] \right)}}{29} = 5$$
$$3 p + 4 q = 29$$
We give the system of equations to the canonical form
$$5 p - 3 q = 0$$
$$3 p + 4 q = 29$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 3 x_{2}\\3 x_{1} + 4 x_{2}\end{matrix}\right] = \left[\begin{matrix}0\\29\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -3\\3 & 4\end{matrix}\right] \right)} = 29$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}0 & -3\\29 & 4\end{matrix}\right] \right)}}{29} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 0\\3 & 29\end{matrix}\right] \right)}}{29} = 5$$
Gaussian elimination
Given the system of equations
$$5 p - 3 q = 0$$
$$3 p + 4 q = 29$$
We give the system of equations to the canonical form
$$5 p - 3 q = 0$$
$$3 p + 4 q = 29$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -3 & 0\\3 & 4 & 29\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -3 & 0\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{3 \cdot 5}{5} & 4 - - \frac{9}{5} & 29 - \frac{0 \cdot 3}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{29}{5} & 29\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -3 & 0\\0 & \frac{29}{5} & 29\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\\frac{29}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{29}{5} & 29\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-15\right) 0}{29} & -3 - \frac{\left(-15\right) 29}{5 \cdot 29} & - \frac{\left(-15\right) 29}{29}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 15\\0 & \frac{29}{5} & 29\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 15 = 0$$
$$\frac{29 x_{2}}{5} - 29 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = 5$$
$$5 p - 3 q = 0$$
$$3 p + 4 q = 29$$
We give the system of equations to the canonical form
$$5 p - 3 q = 0$$
$$3 p + 4 q = 29$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -3 & 0\\3 & 4 & 29\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -3 & 0\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{3 \cdot 5}{5} & 4 - - \frac{9}{5} & 29 - \frac{0 \cdot 3}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{29}{5} & 29\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -3 & 0\\0 & \frac{29}{5} & 29\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\\frac{29}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{29}{5} & 29\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-15\right) 0}{29} & -3 - \frac{\left(-15\right) 29}{5 \cdot 29} & - \frac{\left(-15\right) 29}{29}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 15\\0 & \frac{29}{5} & 29\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 15 = 0$$
$$\frac{29 x_{2}}{5} - 29 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = 5$$