Mister Exam
6x+45y=39; 7x+15y=5
The solution
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6*x + 45*y = 39
$$6 x + 45 y = 39$$
7*x + 15*y = 5
$$7 x + 15 y = 5$$
7*x + 15*y = 5
Detail solution
Given the system of equations
$$6 x + 45 y = 39$$
$$7 x + 15 y = 5$$
Let's express from equation 1 x
$$6 x + 45 y = 39$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$6 x = 39 - 45 y$$
$$6 x = 39 - 45 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{6 x}{6} = \frac{39 - 45 y}{6}$$
$$x = \frac{13}{2} - \frac{15 y}{2}$$
Let's try the obtained element x to 2-th equation
$$7 x + 15 y = 5$$
We get:
$$15 y + 7 \left(\frac{13}{2} - \frac{15 y}{2}\right) = 5$$
$$\frac{91}{2} - \frac{75 y}{2} = 5$$
We move the free summand 91/2 from the left part to the right part performing the sign change
$$- \frac{75 y}{2} = - \frac{91}{2} + 5$$
$$- \frac{75 y}{2} = - \frac{81}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{75}{2} y}{- \frac{75}{2}} = - \frac{81}{\left(- \frac{75}{2}\right) 2}$$
$$y = \frac{27}{25}$$
Because
$$x = \frac{13}{2} - \frac{15 y}{2}$$
then
$$x = \frac{13}{2} - \frac{81}{10}$$
$$x = - \frac{8}{5}$$
The answer:
$$x = - \frac{8}{5}$$
$$y = \frac{27}{25}$$
$$6 x + 45 y = 39$$
$$7 x + 15 y = 5$$
Let's express from equation 1 x
$$6 x + 45 y = 39$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$6 x = 39 - 45 y$$
$$6 x = 39 - 45 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{6 x}{6} = \frac{39 - 45 y}{6}$$
$$x = \frac{13}{2} - \frac{15 y}{2}$$
Let's try the obtained element x to 2-th equation
$$7 x + 15 y = 5$$
We get:
$$15 y + 7 \left(\frac{13}{2} - \frac{15 y}{2}\right) = 5$$
$$\frac{91}{2} - \frac{75 y}{2} = 5$$
We move the free summand 91/2 from the left part to the right part performing the sign change
$$- \frac{75 y}{2} = - \frac{91}{2} + 5$$
$$- \frac{75 y}{2} = - \frac{81}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{75}{2} y}{- \frac{75}{2}} = - \frac{81}{\left(- \frac{75}{2}\right) 2}$$
$$y = \frac{27}{25}$$
Because
$$x = \frac{13}{2} - \frac{15 y}{2}$$
then
$$x = \frac{13}{2} - \frac{81}{10}$$
$$x = - \frac{8}{5}$$
The answer:
$$x = - \frac{8}{5}$$
$$y = \frac{27}{25}$$
Rapid solution
$$x_{1} = - \frac{8}{5}$$
=
$$- \frac{8}{5}$$
=
$$y_{1} = \frac{27}{25}$$
=
$$\frac{27}{25}$$
=
=
$$- \frac{8}{5}$$
=
-1.6
$$y_{1} = \frac{27}{25}$$
=
$$\frac{27}{25}$$
=
1.08000000000000
Cramer's rule
$$6 x + 45 y = 39$$
$$7 x + 15 y = 5$$
We give the system of equations to the canonical form
$$6 x + 45 y = 39$$
$$7 x + 15 y = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}6 x_{1} + 45 x_{2}\\7 x_{1} + 15 x_{2}\end{matrix}\right] = \left[\begin{matrix}39\\5\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}6 & 45\\7 & 15\end{matrix}\right] \right)} = -225$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}39 & 45\\5 & 15\end{matrix}\right] \right)}}{225} = - \frac{8}{5}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}6 & 39\\7 & 5\end{matrix}\right] \right)}}{225} = \frac{27}{25}$$
$$7 x + 15 y = 5$$
We give the system of equations to the canonical form
$$6 x + 45 y = 39$$
$$7 x + 15 y = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}6 x_{1} + 45 x_{2}\\7 x_{1} + 15 x_{2}\end{matrix}\right] = \left[\begin{matrix}39\\5\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}6 & 45\\7 & 15\end{matrix}\right] \right)} = -225$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}39 & 45\\5 & 15\end{matrix}\right] \right)}}{225} = - \frac{8}{5}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}6 & 39\\7 & 5\end{matrix}\right] \right)}}{225} = \frac{27}{25}$$
Gaussian elimination
Given the system of equations
$$6 x + 45 y = 39$$
$$7 x + 15 y = 5$$
We give the system of equations to the canonical form
$$6 x + 45 y = 39$$
$$7 x + 15 y = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}6 & 45 & 39\\7 & 15 & 5\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}6\\7\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}6 & 45 & 39\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{6 \cdot 7}{6} & 15 - \frac{7 \cdot 45}{6} & 5 - \frac{7 \cdot 39}{6}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{75}{2} & - \frac{81}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}6 & 45 & 39\\0 & - \frac{75}{2} & - \frac{81}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}45\\- \frac{75}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{75}{2} & - \frac{81}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{\left(-6\right) 0}{5} & 45 - - -45 & 39 - - \frac{-243}{5}\end{matrix}\right] = \left[\begin{matrix}6 & 0 & - \frac{48}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}6 & 0 & - \frac{48}{5}\\0 & - \frac{75}{2} & - \frac{81}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$6 x_{1} + \frac{48}{5} = 0$$
$$\frac{81}{2} - \frac{75 x_{2}}{2} = 0$$
We get the answer:
$$x_{1} = - \frac{8}{5}$$
$$x_{2} = \frac{27}{25}$$
$$6 x + 45 y = 39$$
$$7 x + 15 y = 5$$
We give the system of equations to the canonical form
$$6 x + 45 y = 39$$
$$7 x + 15 y = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}6 & 45 & 39\\7 & 15 & 5\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}6\\7\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}6 & 45 & 39\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{6 \cdot 7}{6} & 15 - \frac{7 \cdot 45}{6} & 5 - \frac{7 \cdot 39}{6}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{75}{2} & - \frac{81}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}6 & 45 & 39\\0 & - \frac{75}{2} & - \frac{81}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}45\\- \frac{75}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{75}{2} & - \frac{81}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{\left(-6\right) 0}{5} & 45 - - -45 & 39 - - \frac{-243}{5}\end{matrix}\right] = \left[\begin{matrix}6 & 0 & - \frac{48}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}6 & 0 & - \frac{48}{5}\\0 & - \frac{75}{2} & - \frac{81}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$6 x_{1} + \frac{48}{5} = 0$$
$$\frac{81}{2} - \frac{75 x_{2}}{2} = 0$$
We get the answer:
$$x_{1} = - \frac{8}{5}$$
$$x_{2} = \frac{27}{25}$$