Mister Exam
3x+7y=-1; 5x+9y=1
The solution
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3*x + 7*y = -1
$$3 x + 7 y = -1$$
5*x + 9*y = 1
$$5 x + 9 y = 1$$
5*x + 9*y = 1
Detail solution
Given the system of equations
$$3 x + 7 y = -1$$
$$5 x + 9 y = 1$$
Let's express from equation 1 x
$$3 x + 7 y = -1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = - 7 y - 1$$
$$3 x = - 7 y - 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{- 7 y - 1}{3}$$
$$x = - \frac{7 y}{3} - \frac{1}{3}$$
Let's try the obtained element x to 2-th equation
$$5 x + 9 y = 1$$
We get:
$$9 y + 5 \left(- \frac{7 y}{3} - \frac{1}{3}\right) = 1$$
$$- \frac{8 y}{3} - \frac{5}{3} = 1$$
We move the free summand -5/3 from the left part to the right part performing the sign change
$$- \frac{8 y}{3} = 1 + \frac{5}{3}$$
$$- \frac{8 y}{3} = \frac{8}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{8}{3} y}{- \frac{8}{3}} = \frac{8}{\left(- \frac{8}{3}\right) 3}$$
$$y = -1$$
Because
$$x = - \frac{7 y}{3} - \frac{1}{3}$$
then
$$x = - \frac{1}{3} - - \frac{7}{3}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = -1$$
$$3 x + 7 y = -1$$
$$5 x + 9 y = 1$$
Let's express from equation 1 x
$$3 x + 7 y = -1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = - 7 y - 1$$
$$3 x = - 7 y - 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{- 7 y - 1}{3}$$
$$x = - \frac{7 y}{3} - \frac{1}{3}$$
Let's try the obtained element x to 2-th equation
$$5 x + 9 y = 1$$
We get:
$$9 y + 5 \left(- \frac{7 y}{3} - \frac{1}{3}\right) = 1$$
$$- \frac{8 y}{3} - \frac{5}{3} = 1$$
We move the free summand -5/3 from the left part to the right part performing the sign change
$$- \frac{8 y}{3} = 1 + \frac{5}{3}$$
$$- \frac{8 y}{3} = \frac{8}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{8}{3} y}{- \frac{8}{3}} = \frac{8}{\left(- \frac{8}{3}\right) 3}$$
$$y = -1$$
Because
$$x = - \frac{7 y}{3} - \frac{1}{3}$$
then
$$x = - \frac{1}{3} - - \frac{7}{3}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = -1$$
Rapid solution
$$x_{1} = 2$$
=
$$2$$
=
$$y_{1} = -1$$
=
$$-1$$
=
=
$$2$$
=
2
$$y_{1} = -1$$
=
$$-1$$
=
-1
Cramer's rule
$$3 x + 7 y = -1$$
$$5 x + 9 y = 1$$
We give the system of equations to the canonical form
$$3 x + 7 y = -1$$
$$5 x + 9 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + 7 x_{2}\\5 x_{1} + 9 x_{2}\end{matrix}\right] = \left[\begin{matrix}-1\\1\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 7\\5 & 9\end{matrix}\right] \right)} = -8$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-1 & 7\\1 & 9\end{matrix}\right] \right)}}{8} = 2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -1\\5 & 1\end{matrix}\right] \right)}}{8} = -1$$
$$5 x + 9 y = 1$$
We give the system of equations to the canonical form
$$3 x + 7 y = -1$$
$$5 x + 9 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + 7 x_{2}\\5 x_{1} + 9 x_{2}\end{matrix}\right] = \left[\begin{matrix}-1\\1\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 7\\5 & 9\end{matrix}\right] \right)} = -8$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-1 & 7\\1 & 9\end{matrix}\right] \right)}}{8} = 2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -1\\5 & 1\end{matrix}\right] \right)}}{8} = -1$$
Gaussian elimination
Given the system of equations
$$3 x + 7 y = -1$$
$$5 x + 9 y = 1$$
We give the system of equations to the canonical form
$$3 x + 7 y = -1$$
$$5 x + 9 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 7 & -1\\5 & 9 & 1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 7 & -1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{3 \cdot 5}{3} & 9 - \frac{5 \cdot 7}{3} & 1 - - \frac{5}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{8}{3} & \frac{8}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 7 & -1\\0 & - \frac{8}{3} & \frac{8}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}7\\- \frac{8}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{8}{3} & \frac{8}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-21\right) 0}{8} & 7 - - -7 & -1 - \frac{\left(-21\right) 8}{3 \cdot 8}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 6\\0 & - \frac{8}{3} & \frac{8}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 6 = 0$$
$$- \frac{8 x_{2}}{3} - \frac{8}{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = -1$$
$$3 x + 7 y = -1$$
$$5 x + 9 y = 1$$
We give the system of equations to the canonical form
$$3 x + 7 y = -1$$
$$5 x + 9 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 7 & -1\\5 & 9 & 1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 7 & -1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{3 \cdot 5}{3} & 9 - \frac{5 \cdot 7}{3} & 1 - - \frac{5}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{8}{3} & \frac{8}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 7 & -1\\0 & - \frac{8}{3} & \frac{8}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}7\\- \frac{8}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{8}{3} & \frac{8}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-21\right) 0}{8} & 7 - - -7 & -1 - \frac{\left(-21\right) 8}{3 \cdot 8}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 6\\0 & - \frac{8}{3} & \frac{8}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 6 = 0$$
$$- \frac{8 x_{2}}{3} - \frac{8}{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = -1$$